Download Chapter 4: Quantities of Reactants and Products

Chapter 4: Quantities of Reactants and Products
121
Chapter 4: Quantities of Reactants and Products
Review Questions
1.
A balanced chemical equation indicates the relative amounts of reactants and products so that the number
of atoms of each element in the reactants equals the number of atoms of the same elements in the products.
2.
Define the problem: Given the equation for a reaction, find the related moles, molecules, and masses of the
reactants and products.
Develop a plan: The stoichiometric coefficients (the numbers in front of the molecules’ formulas) in the
balanced equation can be interpreted as the related number of moles or the related number of molecules.
The molar mass of each molecule is multiplied by the number of moles of that molecule represented in the
equation to find its mass in grams.
Execute the plan: 3 H2(g) + N2(g)
2 NH3(g)
The stoichiometric coefficients are 3 for H2, 1 for N2, and 2 for NH3. (Note: Molar masses need only be
precise enough to give two decimal places in the calculated masses, comparable to the precision in the given
mass of NH3.)
H2
N2
NH3(g)
3 mol
1 mol
2 mol
3 molecules
1 molecule
2 molecules
3 mol × (2.02g/mol) = 6.06 g
1 mol × (28.02g/mol) = 28.02 g
34.08 g
Check your answers: The conservation of mass says that the masses of the reactants must add up to the
masses of the products. 6.06 g + 28.02 g = 34.08 g. This looks right.
3.
The absolute quantity of each of the reactants is variable and ultimately controlled by the experimenter. The
reactants could be present in any relative quantities, though they only react with each other according to
the balanced chemical equation. In most cases, that means that one or more of the reactants will be left over
when the reaction stops, since it eventually runs out of at least one of the reactants. If, however, the
relative quantities of reactants are identical to the relative quantities required by the balanced chemical
equation, then the reactants will all run out simultaneously. This special condition is noted by saying that
“the reactants were present in stoichiometric amounts.”
4.
Define the problem: Given the equation for a reaction, write all the stoichiometric factors represented by it.
Develop a plan: The stoichiometric factors are ratios describing two related quantities. Take each
component of the reaction and relate it one by one to the other three components of the reaction, then put
each pair-wise relationship in unit factor form.
Execute the plan: 3 MgO(s) + 2 Fe(s)
Fe2O3(s) + 3 Mg(s)
3 mol MgO reacts with 2 mol Fe to form 1 mol Fe2O3 and 3 moles Mg.
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Chapter 4: Quantities of Reactants and Products
3 mol MgO
,
1 mol Fe2 O3
3 mol MgO
,
2 mol Fe
3 mol MgO
1 mol MgO
or
3 mol Mg
1 mol Mg
2 mol Fe reacts with 3 mol MgO to form 1 mol Fe2O3 and 3 moles Mg.
2 mol Fe
,
1 mol Fe2 O3
2 mol Fe
,
3 mol MgO
2 mol Fe
3 mol Mg
1 mol Fe2O3 is formed with 3 moles Mg when 2 mol Fe reacts with 3 mol MgO.
1 mol Fe2 O3
,
3 mol Mg
1 mol Fe2 O3
,
2 mol Fe
1 mol Fe2 O3
3 mol MgO
3 moles Mg are formed with 1 mol Fe2O3 when 2 mol Fe reacts with 3 mol MgO.
3 mol Mg
,
1 mol Fe2 O3
3 mol Mg
,
2 mol Fe
3 mol Mg
1 mol Mg
or
3 mol MgO
1 mol MgO
Check your answers: With four different chemicals in this equation we can have 4 × 4 pairs. Since it is not
important to relate a molecule to itself, that reduces the number of different pairs by 4. That means we
should have (4 × 4) – 4 = 12 stoichiometric factors, and that’s how many we have.
5.
If reactants are combined in stoichiometric quantities, they will both run out at the same time. Therefore, all
their atoms and the sum of their masses will be present in the lone product, CO2, according to the law of
conservation of mass.
10.0 g + 26.6 g = 36.6 g CO2 is produced.
Check your answer: Though it is not necessary, we could check this by using the molar mass to calculate
the moles of C or O2 used. Then use the stoichiometry of the balanced equation (C + O2
CO2) to
find the moles of product. Then find the grams using the molar mass of CO2.
10.0 g C ×
6.
1 mol C
1 mol CO 2 44.0095 g CO2
×
×
= 36.6 g CO2
12.0107 g C
1 mol C
1 mol CO2
Define the problem: Given a reaction, determine the conversion factors needed to convert between grams
and mole, grams and grams, and moles and grams.
Develop a plan: The molar mass and stoichiometric coefficients can be used for the appropriate unit
factors.
Execute the plan: 2 Fe(s) + 3 Cl2(g)
2 FeCl3(s)
To take the grams of Cl2 to moles of Cl2, use the molar mass of Cl2:
g Cl2 ×
1 mol Cl 2
= mol Cl2
70.906 g Cl 2
To take the moles of Cl2 to moles FeCl3, use the stoichiometry of the balanced equation:
mol Cl2 ×
2 mol FeCl 3
= mol FeCl3
3 mol Cl 2
Chapter 4: Quantities of Reactants and Products
123
To take the grams of FeCl3 to moles of FeCl3, use the molar mass of FeCl3:
g FeCl3 ×
1 mol FeCl 3
= mol FeCl3
162.204 g FeCl 3
To take the grams of Cl2 to grams of FeCl3, combine all three factors together:
1 mol Cl 2
2 mol FeCl 3 162.204 g FeCl 3 324.40 g FeCl 3
×
×
=
70.904 g Cl 2
3 mol Cl 2
1 mol FeCl 3
212.7 g Cl 2
g Cl2 ×
324.408 g FeCl 3
= g FeCl3
212.718 g Cl 2
Hence, the scheme looks like this:
g Cl 2
324.408 g FeCl 3
212.718 g Cl 2
1 mol Cl 2
70.906 g Cl 2
mol Cl 2
g FeCl 3
1 mol FeCl 3
162.204 g FeCl 3
2 mol FeCl 3
3 mol Cl 2
mol FeCl 3
Check your answers: Going either way around the circuit, the results end up the same.
7.
The limiting reactant is the reactant that runs out first. The reaction’s stoichiometry is based on how many
reactant molecules are present, not how heavy they are. There are two different reasons why the low mass
reactant may not be the limiting reactant. First, the reactant with the smallest mass may be a lightweight
element, such as hydrogen. With a low molar mass, the small given mass might still represent a relatively
large number of molecules. Second, if the low mass reactant has a relatively small stoichiometric coefficient,
then other reactants must be present in larger proportions. Therefore, there is no guarantee that the reactant
present with the smallest mass is going to end up being the limiting reactant.
8.
A limiting reactant needs to be a reactant. The products are produced in the reaction; therefore they do not
qualify.
9.
The limiting reactant determines the theoretical yield. If nothing goes wrong and all the reactants that can
react do react, then the quantity of limiting reactant should prescribe the quantity of product that will form.
The actual yield will only be related to the limiting reactant in that it represents an upper limit to that
quantity.
Stoichiometry
10. Define the problem: Given the equation for the reaction, find the related molecules, atoms, moles, masses of
each of the reactants and products, and total masses of the reactants and the products.
Develop a plan: The stoichiometric coefficients (the numbers in front of the molecules’ formulas) in the
balanced equation can be interpreted as the related number of molecules or the related number of moles of
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Chapter 4: Quantities of Reactants and Products
molecules. The molar mass of each molecule is multiplied by the number of moles of that molecule
represented in the equation to find its mass in grams. Add up the mass of KOH and HCl for the total
reactant mass. Add up the mass of KCl and H2O for the total product mass.
Execute the plan: KOH(aq) + HCl(aq)
KCl(aq) + 6 H2O(l)
The stoichiometric coefficients are 1 for KOH(aq), 1 for HCl(aq), 1 for KCl(aq) and 1 for H2O(l).
No. molecules
No. atoms
No. moles of
molecules
Mass
KOH
HCl
KCl
H2O
1
1
1
1
1 K, 1 O and 1 H
1 H and 1 Cl
1 K and 1 Cl
2 H and 1 O
1
1
1
1
1 mol ×
(56.1056 g/mol) =
56.1056 g
1 mol ×
(36.4609 g/mol) =
224.00 g
1 mol ×
(74.5513 g/mol) =
176.04 g
1 mol ×
(18.0152 g/mol) =
108.12 g
Total mass of
reactants
92.5665 g
Total mass of
products
–
–
92.5665 g
Check your answers: A properly balanced equation has the same numbers of atoms of each type (1K, 1 O, 1
Cl, and 2 H) in the products and reactants. The law of conservation of mass says that masses of the
reactants must add up to the masses of the products, 92.5665 g.
11. Define the problem: Given the equation for the reaction, find the related molecules, atoms, moles, masses of
each of the reactants and products, and total masses of the reactants and t he products.
Develop a plan: The stoichiometric coefficients (the numbers in front of the molecules’ formulas) in the
balanced equation can be interpreted as the related number of molecules or the related number of moles of
molecules. The molar mass of each molecule is multiplied by the number of moles of that molecule
represented in the equation to find its mass in grams. Add up the mass of C2H6 and O2 for the total reactant
mass. Add up the mass of CO2 and H2O for the total product mass.
Execute the plan: 2 C2H6(g) + 7 O2(g)
4 CO2(g) + 6 H2O(g)
The stoichiometric coefficients are 2 for C2H6, 7 for O2, 4 for CO2, and 6 for H2O.
No. molecules
No. atoms
No. moles of
molecules
Mass
C2H6
O2
CO2
H2O
2
7
4
6
4 C and 12 H
14 O
4 C and 8 O
12 H and 6 O
2
7
4
6
2 mol ×
7 mol ×
4 mol ×
6 mol ×
Chapter 4: Quantities of Reactants and Products
(30.0688 g/mol) =
60.16 g
Total mass of
reactants
(31.9988 g/mol) =
224.00 g
(44.0095 g/mol) =
176.04 g
284.16 g
Total mass of
products
125
(18.0152 g/mol) =
108.12 g
–
–
284.16 g
Check your answers: A properly balanced equation has the same numbers of atoms of each type (4 C, 12 H,
and 14 O) in the products and reactants. The law of conservation of mass says that masses of the reactants
must add up to the masses of the products, 284.16 g. This looks right.
12. Define the problem: Given the balanced equation for a reaction, identify the stoichiometric coefficients in
this equation, and relate the quantity of products to reactants and vice versa.
Develop a plan: (a) The law of conservation of mass says that the total mass of the reactants is the same as
the total mass of the products. (b) The stoichiometric coefficients are the numbers in front of each formula in
the equation. (c) The stoichiometric coefficients can be interpreted as the related number of reactants. Use
the formula stoichiometry of O2 to find the number of molecules that reacted. Use equation stoichiometry to
find out how many atoms of Mg reacted with that many O2 molecules.
Execute the plan: First, balance the equation:
2 Mg(s) + O2(g)
2 MgO(g)
(a) The total mass of product, MgO, is 1.00 g. The total mass of the reactants (mass of Mg plus mass of O2)
that reacted must also be 1.00 g, due to the conservation of mass.
(b) The stoichiometric coefficients are 2 for Mg, 1 for O2, and 2 for MgO.
(c) 50 atoms of oxygen make up 25 molecules of O2, since there are two O atoms in each molecule. Since
the stoichiometry is 1:1, that means 50 atoms of Mg were needed to react with this much oxygen.
Check your answers: Though it is an unnecessary calculation, we can check the answer to part (a) by
calculating the masses of Mg and O2 that reacted to make 1.00 gram of MgO,
1.00 g MgO×
1 mol MgO
1 mol Mg 24.3050 g Mg
×
×
= 0.603 g Mg
40.3044 g MgO 1 mol MgO
1 mol Mg
1.00 g MgO×
1 mol MgO
1 mol O2 31.9988 g O 2
×
×
= 0.397 g O2
40.3044 g MgO 2mol MgO
1 mol O 2
then adding them up: 0.603 g + 0.397 g = 1.000 g. This answer looks right.
13. Define the problem: Given the balanced equation for a reaction and the mass of one reactant, determine
how may grams of a product will be formed and how many grams of the other reactant will be required.
Develop a plan: The stoichiometric coefficients can be interpreted as the related number of moles of
reactants and products. Use the molar masses to convert mass to moles and the equation stoichiometry to
relate moles.
Execute the plan:
1.0 g C12 H22 O11 ×
1 mol C12 H 22O11
12 mol CO2
44.0095 g CO2
×
×
= 1.5 g CO2
342.2956 g C 12H22O11 1 mol C12H 22O11
1 mol CO2
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Chapter 4: Quantities of Reactants and Products
1.0 g C12 H22 O11 ×
1 mol C12 H 22O11
12 mol O 2
31.9988 g O2
×
×
= 1.1 g O2
342.2956 g C 12H22O11 1 mol C 12H 22O11
1 mol O2
Check your answers: The sum of the reactants masses should equal the sum of the product masses:
1.0 g C12 H22 O11 ×
1 mol C12 H 22O11
11 mol H2O
18.0152 g H2O
×
×
= 0.58 g H2O
342.2956 g C 12H22O11 1 mol C12H 22O11
1 mol H2 O
Adding them: 1.0 g + 1.1 g = 2.1 g, and 1.5 g + 0.58 g = 2.1 g.
Chapter 4: Quantities of Reactants and Products
127
14. Define the problem: Write a balanced equation for a combination reaction.
Develop a plan: Use appropriate numbers in front of each reactant and product (stoichiometric coefficients)
to balance the equation.
Execute the plan: 4 Fe(s) + 3 O2(g)
2 Fe2O3(s)
Check your answer: 4 Fe & 6 O
15. In the first box, three A 2 molecules are combined with six B atoms. After the reaction occurs, the box
contains only six AB molecules. That means symbolically:
3 A2 + 6 B
6 AB
Therefore, the equation given in (b), A 2 + 2 B
between reactants and products.
16.
2 AB, shows the right stoichiometric relationship
2 × 4 A2 + 2 × 3 B
2 × 1 B3A 8
8 A2 + 6 B
2 B3A 8
17. Balance Cl atoms (put a “3” in front of Cl2 and a “2” in front of SbCl3, so each side of the equation has 6 Cl
atoms, 3 × 2 = 2 × 3 = 6). Balance Sb atoms (put a 2 in front of Sb, so each side of the equation has 2 Sb
atoms): 2 Sb + 3 Cl2
2 SbCl3.
Box (a) represents the correct reactants. Box (c) represents the correct products.
Classification of Chemical Reactions
18. Define the problem: Write the balanced formation equations.
Develop a plan: Write the formula of the compound as the product, and write the reactants in the form of
elements. Note, some elements are in the form of diatomic molecules (e.g., O2). Balance the equations.
Execute the plan:
(a) Carbon monoxide = CO(g).
(b) Nickel(II) oxide = NiO(s).
2 C(s) + O2(g)
2 Ni(s) + O2(g)
(c) Chromium(III) oxide = Cr2O3(s).
2 CO(g)
2 NiO(s)
4 Cr(s) + 3 O2(g)
Check your answers: (a) 2 C & 2 O (b) 2 Ni & 2 O (c) 4 Cr & 6 O
2 Cr2O3(s)
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Chapter 4: Quantities of Reactants and Products
19. Define the problem: Write the balanced formation equations.
Develop a plan: Write the formula of the compound as the product, and write the reactants in the form of
elements. Note, some elements are in the form of diatomic molecules (e.g., O2). Balance the equations.
Execute the plan:
(a) Copper(I) oxide = Cu 2O(s).
4 Cu(s) + O2(g)
(b) Arsenic(III) oxide = As 2O3(s).
(c) Zinc oxide = ZnO(s).
2 Cu 2O(s)
4 As(s) + 3 O2(g)
2 Zn(s) + O2(g)
2 As2O3(s)
2 ZnO(s)
Check your answers: (a) 4 Cu & 2 O (b) 4 As & 6 O (c) 2 Zn & 2 O
20. Define the problem: Write balanced decomposition equations for some carbonate salts.
Develop a plan: Assume that the products of decomposition are oxides and carbon dioxide as described in
the text. Write the formula of the compound as the reactant and determine the formula of the metal oxide.
Write the products in the form of the metal oxide and carbon dioxide. Balance the equations.
Execute the plan:
(a) BeCO3(s)
BeO(s) + CO2(g) beryllium oxide, carbon dioxide
(b) NiCO3(s)
NiO(s) + CO2(g) nickel(II) oxide, carbon dioxide
(c) Al2(CO3)3(s)
Al2O3(s) + 3 CO2(g)
aluminum oxide, carbon dioxide
Check your answers: (a) 1 Be, 1 C, & 3 O (b) 1 Ni, 1 C, & 3 O (c) 2 Al, 3 C & 9 O
21. Define the problem: Assume that the products of decomposition are oxides and carbon dioxide as
described in the text. Write the formula of the compound as the reactant and determine the formula of the
metal oxide. Write the products in the form of the metal oxide and carbon dioxide. Balance the equations.
Execute the plan:
(a) ZnCO3(s)
ZnO(s) + CO2(g)
zinc oxide, carbon dioxide
(b) MnCO3(s)
MnO(s) + CO2(g)
manganese(II) oxide, carbon dioxide
(c) PbCO3(s)
PbO(s) + CO2(g)
lead(II) oxide, carbon dioxide
Check your answers: (a) 1 Zn, 1 C, & 3 O (b) 1 Mn, 1 C, & 3 O (c) 1 Pb, 1 C, & 3 O
22. Define the problem: Write the balanced combustion equations.
Develop a plan: The products of combustion depend on the elements present in the compound being
combusted:
Compound Contains:
Combustion Product is:
C
H
CO2
H2O
Chapter 4: Quantities of Reactants and Products
129
The compound and O2 are the reactant. The products are CO2 and H2O for these combustion reactions.
Balance the equations, starting with C, then H, then O. If the coefficient of O2 ends up a fraction, multiply
everything in the reaction by two, to make all coefficients whole numbers.
Execute the plan:
(a) Unbalanced:
? C4H10(g) + ? O2(g)
? CO2(g) + ? H2O(g)
Balance C:
C4H10(g) + ? O2(g)
4 CO2(g) + ? H2O(g)
4 C’s
Balance H:
C4H10(g) + ? O2(g)
4 CO2(g) + 5 H2O(g)
10 H’s
Balance O:
C4H10(g) +
Multiply by 2:
2 C4H10(g) + 13 O2(g)
(b) Unbalanced:
13
O2(g)
2
4 CO2(g) + 5 H2O(g)
13 O’s
8 CO2(g) + 10 H2O(g)
? C6H12O6(s) + ? O2(g)
? CO2(g) + ? H2O(g)
Balance C:
C6H12O6(s) + ? O2(g)
6 CO2(g) + ? H2O(g)
6 C’s
Balance H:
C6H12O6(s) + ? O2(g)
6 CO2(g) + 6 H2O(g)
12 H’s
Balance O:
C6H12O6(s) + 6 O2(g)
6 CO2(g) + 6 H2O(g)
18 O’s
? C4H8O(l) + ? O2(g)
? CO2(g) + ? H2O(g)
(c) Unbalanced:
Balance C:
C4H8O(l) + ? O2(g)
4 CO2(g) + ? H2O(g)
4 C’s
Balance H:
C4H8O(l) + ? O2(g)
4 CO2(g) + 4 H2O(g)
8 H’s
Balance O:
C4H8O(l) +
Multiply by 2:
2 C4H8O(l) + 11 O2(g)
11
O2(g)
2
4 CO2(g) + 4 H2O(g)
12 O’s
8 CO2(g) + 8 H2O(g)
Check your answers: (a) 8 C, 20 H, & 26 O (b) 6 C, 12 H, & 18 O (c) 8 C, 16 H, & 24 O
23. Define the problem: Balance the equation for the reaction between a metal and oxygen.
Develop a plan: The products are metal oxides. Write the product in the form of a neutral metal oxide using
the periodic table to predict the metal ion’s charge. Balance the equations. Note that the oxide anion has a
–2 charge.
Execute the plan: (a) Magnesium is in Group 2A. Its cation will have +2 charge.
2 Mg(s) + O2(g)
2 MgO(s) magnesium oxide
(b) Calcium is in Group 2A. Its cation will have +2 charge.
2 Ca(s) + O2(g)
2 CaO(s)
calcium oxide
(c) Indium is in group 3A. Its cation will have +3 charge.
4 In(s) + 3 O2(g)
2 In 2O3(s) indium(III) oxide
Check your answers: (a) 2 Mg & 2 O (b) 2 Ca & 2 O (c) 4 In & 6 O
130
Chapter 4: Quantities of Reactants and Products
24. Define the problem: Balance the equation for the reaction between a metal and oxygen.
Develop a plan: Write the product formula in the form of a neutral oxide. Note that the Roman numeral in a
metal oxide’s name is used to get the metal ion’s charge. Balance the equations. Note that the oxide anion
has a –2 charge.
Execute the plan:
(a) Titanium(IV) oxide, contains the titanium(IV) cation with a +4 charge: TiO2
Ti(s) + O2(g)
TiO2(s)
(b) Sulfur dioxide is SO2.
S8(s) + 8 O2(g)
(c) Selenium dioxide is SeO2.
Se(s) + O2(g)
8 SO2(g)
SeO2(s)
Check your answers: (a) 1 Ti & 2 O (b) 8 S & 16 O (c) 1 Se & 2 O
25. Define the problem: Balance the equation for the reaction between a metal and a halogen.
Develop a plan: Write the product formula in the form of a neutral halide salt. Balance the equations. Note
that the halogens form anions with a –1 charge.
Execute the plan:
(a) Potassium is in group 1A. Its cation will have +1 charge.
2 K(s) + Cl2(g)
2 KCl(s)
potassium chloride
(b) Magnesium is in group 2A. Its cation will have +2 charge.
Mg(s) + Br2(l)
MgBr2(s)
magnesium bromide
(c) Aluminum is in group 3A. Its cation will have +3 charge.
2 Al(s) + 3 F2(g)
2 AlF3(s) aluminum fluoride
Check your answers: (a) 2 K & 2 Cl (b) 1 Mg & 2 Br (c) 2 Al & 6 F
26. Define the problem: Balance the equation for reaction between a metal and a halogen.
Develop a plan: Write the product formula in the form of a neutral halide salt. Note that the Roman numeral
in a metal oxide’s name is used to get the metal ion’s charge. Balance the equations. Note that the halogens
form anions with a –1 charge.
Execute the plan:
(a) Chromium(III) has a +3 charge, so chromium(III) chloride is: CrCl3
2 Cr(s) + 3 Cl2(g)
2 CrCl3(s)
(b) Copper(II) has a +2 charge, so copper(II) bromide is: CuBr2
Cu(s) + Br2(l)
CuBr2(s)
(c) Platinum(IV) has a +4 charge, so platinum(IV) fluoride is: PtF4
Pt(s) + 2 F2(g)
PtF4(s)
Chapter 4: Quantities of Reactants and Products
131
Check your answers: (a) 2 Cr & 6 Cl (b) 1 Cu & 2 Br (c) 1 Pt & 4 F
Balancing Equations
27. Define the problem: Balance the given equations.
Develop a plan: When balancing equations, select a specific order in which the elements are balanced.
Select first the atoms that are only in one product and one reactant, since they will be easier. Select next
those elements that are in more than one reactant or product (such as H or O) and then those which are
present in elemental form in either the reactants or products. Be systematic. If any of the coefficients end
up fractional, multiply every coefficient by the same constant to eliminate the fraction.
Execute the plan:
(a) ? Al(s) + ? O2(g)
? Al2O3(s)
2 Al(s) + ? O2(g)
1 Al2O3(s)
2 Al(s) +
3
O2(g)
2
4 Al(s) + 3 O2(g)
(b) ? N2(g) + ? H2(g)
2 Al
1 Al2O3(s)
3O
2 Al2O3(s)
? NH3(g) Select order: H then N
? N2(g) + 3 H2(g)
2 NH3(g) 6 H
1 N2(g) + 3 H2(g)
2 NH3(g) 2 N
(c) ? C6H6(l) + ? O2(g)
Select order: Al then O
? H2O(l) + ? CO2(g)
Select order: C, H then O
C6H6(l) + ? O2(g)
? H2O(l) + 6 CO2(g)
6C
C6H6(l) + ? O2(g)
3 H2O(l) + 6 CO2(g)
6H
C6H6(l) +
15
O2(g)
2
2 C6H6(l) +15 O2(g)
3 H2O(l) + 6 CO2(g)
15 O
6 H2O(l) + 12 CO2(g)
Check your answers: (a) 4 Al & 6 O (b) 2 N & 6 H (c) 12 C, 12 H & 30 O
28. Define the problem: Balance the given equations.
Develop a plan: When balancing equations, select a specific order in which the elements are balanced.
Select first the atoms that are only in one product and one reactant, since they will be easier. Select next
those elements that are in more than one reactant or product (such as H or O) and then those which are
present in elemental form in either the reactants or products. Be systematic. If any of the coefficients end
up fractional, multiply every coefficient by the same constant to eliminate the fraction.
Execute the plan:
(a) ? Fe(s) + ? Cl2(g)
? FeCl3(s)
? Fe(s) + 3 Cl2(g)
2 FeCl3(s)
Select order: Cl then Fe
6 Cl
132
Chapter 4: Quantities of Reactants and Products
2 Fe(s) + 3 Cl2(g)
2 FeCl3(s)
2 Fe
Chapter 4: Quantities of Reactants and Products
(b) ? SiO2(s) + ? C(s)
? Si(s) + ? CO(g)
Select order: O, Si then C
SiO2(s) + ? C(s)
? Si(s) + 2 CO(g)
2O
SiO2(s) + ? C(s)
Si(s) + 2 CO(g)
1 Si
SiO2(s) + 2 C(s)
Si(s) + 2 CO(g)
2C
(c) ? Fe(s) + ? H2O(g)
? Fe3O4(s) + ? H2(g)
Select order: O, Fe then H
? Fe(s) + 4 H2O(g)
Fe3O4(s) + ? H2(g)
4O
3 Fe(s) + 4 H2O(g)
Fe3O4(s) + ? H2(g)
3 Fe
3 Fe(s) + 4 H2O(g)
Fe3O4(s) + 4 H2(g)
8H
Check your answers: (a) 2 Fe & 6 Cl (b) 1 Si, 2 O & 2 C (c) 3 Fe, 8 H & 4 O
29. Define the problem: Balance the given equations.
Develop a plan: Follow the method described in the answers to Question 28.
Execute the plan:
(a) ? UO2(s) + ? HF(l)
? UF4(s)+ ? H2O(l)
Select order: U, F, H, O
UO2(s) + ? HF(l)
UF4(s)+ ? H2O(l)
1U
UO2(s) + 4 HF(l)
UF4(s)+ ? H2O(l)
4F
UO2(s) + 4 HF(l)
UF4(s)+ 2 H2O(l)
4 H and 2 O
(b) ? B2O3(s) + ? HF(l)
? BF3(s) + ? H2O(l)
Select order: B, F, H, O
B2O3(s) + ? HF(l)
2 BF3(s) + ? H2O(l)
2B
B2O3(s) + 6 HF(l)
2 BF3(s) + ? H2O(l)
6F
B2O3(s) + 6 HF(l)
2 BF3(s) + 3 H2O(l)
6 H and 3 O
(c) ? BF3(g) + ? H2O(l)
? HF(g) + ? H3BO3(s)
Select order: B, F, H, O
BF3(g) + ? H2O(l)
? HF(l) + H3BO3(s)
1B
BF3(g) + ? H2O(l)
3 HF(l) + H3BO3(s)
3F
BF3(g) + 3 H2O(l)
3 HF(l) + H3BO3(s)
6 H and 3 O
Check your answers: (a) 1 U, 2 O, 4 H & 4 F (b) 2 B, 3 O, 6 H & 6 F (c) 1 B, 3 F, 6 H & 3 O
30. Define the problem: Balance the given equations.
Develop a plan: Follow the method described in the answers to Question 29.
Execute the plan:
(a) ? MgO(s) + ? Fe(s)
? Fe2O3(s) + ? Mg(s)
Select order: O, Mg, Fe
133
134
Chapter 4: Quantities of Reactants and Products
3 MgO(s) + ? Fe(s)
Fe2O3(s) + ? Mg(s)
3O
3 MgO(s) + ? Fe(s)
Fe2O3(s) + 3 Mg(s)
3 Mg
3 MgO(s) + 2 Fe(s)
Fe2O3(s) + 3 Mg(s)
2 Fe
(b) ? H3BO3(s)
? B2O3(s) + ? H2O(l)
Select order: B, H, O
2 H3BO3(s)
B2O3(s) + ? H2O(l)
2B
2 H3BO3(s)
B2O3(s) + 3 H2O(l)
H and 6 O
(c) ? NaNO3(s) + ? H2SO4(aq)
? Na 2SO4(aq) + ? HNO3(g)
+
+
2–
+
2 NaNO3(s) + ? H2SO4(aq)
Na 2SO4(aq) + ? HNO3(g)
2 Na
2 NaNO3(s) + ? H2SO4(aq)
Na 2SO4(aq) + 2 HNO3(g)
2 NO3
2 NaNO3(s) + H2SO4(aq)
–
Select order: Na , NO3 , H , SO4
Na 2SO4(aq) + 2 HNO3(g)
–
+
2–
2 H and 1 SO4
Check your answers: (a) 3 Mg, 3 O & 2 Fe, (b) 6 H, 2 B & 6 O, (c) 2 Na, 2 N, 10 O, 2 H & 1 S
31. Define the problem: Balance the given equations.
Develop a plan: Follow the method described in the answers to Question 28.
Execute the plan:
(a) ? H2NCl(aq) + ? NH3(g)
? NH4Cl(aq) + ? N2H4(aq)
Select order: Cl, N, H
H2NCl(aq) + ? NH3(g)
NH4Cl(aq) + ? N2H4(aq)
1 Cl
H2NCl(aq) + 2 NH3(g)
NH4Cl(aq) + N2H4(aq)
3 N and 8 H
(b) ? (CH3)2N2H2(l) + ? N2O4(g)
? N2(g) + ? H2O(l) + ? CO2(g)
Select order: H, C, O, N
(CH3)2N2H2(l) + ? N2O4(g)
? N2(g) + 4 H2O(g) + ? CO2(g)
8H
(CH3)2N2H2(l) + ? N2O4(g)
? N2(g) + 4 H2O(g) + 2 CO2(g)
2C
(CH3)2N2H2(l) + 2 N2O4(g)
? N2(g) + 4 H2O(g) + 2 CO2(g)
8O
(CH3)2N2H2(l) + 2 N2O4(g)
3 N2(g) + 4 H2O(g) + 2 CO2(g)
6N
(c) ? CaC2(s) + ? H2O(l)
? Ca(OH)2(s) + ? C2H2(g)
Select order: Ca, C, O, H
CaC2(s) + ? H2O(l)
Ca(OH)2(s) + ? C2H2(g)
1 Ca
CaC2(s) + ? H2O(l)
Ca(OH)2(s) + C2H2(g)
2C
CaC2(s) + 2 H2O(l)
Ca(OH)2(s) + C2H2(g)
2 O and 4 H
Check your answers: (a) 8 H, 3 N & 1 Cl (b) 2 C, 8 H, 6 N & 8 O (c) 1 Ca, 2 C, 4 H & 2 O
Chapter 4: Quantities of Reactants and Products
135
32. Define the problem: Balance the given equations.
Develop a plan: Follow the method described in the answers to Question 29.
Execute the plan:
(a) ? CaNCN(s) + ? H2O(l)
? CaCO3(s) + ? NH3(g) Select order: Ca, N, C, O, H
CaNCN(s) + ? H2O(l)
CaCO3(s) + ? NH3(g)
1 Ca
CaNCN(s) + ? H2O(l)
CaCO3(s) + 2 NH3(g)
2 N and 1 C
CaNCN(s) + 3 H2O(l)
CaCO3(s) + 2 NH3(g)
3 O and 6 H
(b) ? NaBH4(s) + ? H2SO4(aq)
? B2H6(g) + ? H2(g) + ? Na 2SO4(g)
Select order: B, Na, S, O, H
2 NaBH4(s) + ? H2SO4(aq)
B2H6(g) + ? H2(g) + ? Na 2SO4(aq)
2B
2 NaBH4(s) + ? H2SO4(aq)
B2H6(g) + ? H2(g) + Na 2SO4(aq)
2 Na
2 NaBH4(s) + H2SO4(aq)
B2H6(g) + ? H2(g) + Na 2SO4(aq)
1 S and 4 O
2 NaBH4(s) + H2SO4(aq)
B2H6(g) + 2 H2(g) + Na 2SO4(aq)
10 H
(c) ? H2S(aq) + ? Cl2(aq)
? S8(s) + ? HCl(aq)
Select order: S, H, Cl
8 H2S(aq) + ? Cl2(aq)
S8(s) + ? HCl(aq)
8S
8 H2S(aq) + ? Cl2(aq)
S8(s) + 16 HCl(aq)
16 H
8 H2S(aq) + 8 Cl2(aq)
S8(s) + 16 HCl(aq)
16 Cl
Check your answers:
(a) 1 Ca, 2 N, 6 H & 3 O, (b) 2 Na, 2 B, 10 H, 1 S & 4 O
(c) 16 H, 8 S & 16 Cl
33. Define the problem: Balance the given combustion equations.
Develop a plan: When balancing combustion equations, select a C, H, O as the order in which the elements
are balanced. Be systematic. If the coefficient of O2 ends up a fraction, multiply every coefficient by 2 to
make all of the coefficients whole numbers.
Execute the plan:
(a) ? C6H12O6 + ? O2
? CO2 + ? H2O
C6H12O6 + ? O2
6 CO2 + ? H2O
6C
C6H12O6 + ? O2
6 CO2 + 6 H2O
12 H
C6H12O6 + 6 O2
6 CO2 + 6 H2O
18 O
(b) ? C5H12 + ? O2
? CO2 + ? H2O
C5H12 + ? O2
5 CO2 + ? H2O
5C
C5H12 + ? O2
5 CO2 + 6 H2O
12 H
136
Chapter 4: Quantities of Reactants and Products
C5H12 + 8 O2
(c) ? C7H14 + ? O2
5 CO2 + 6 H2O
16 O
? CO2 + ? H2O
C7H14O2 + ? O2
7 CO2 + ? H2O
7C
C7H14O2 + ? O2
7 CO2 + 7 H2O
14 H
C7H14O2 +
19
O2
2
7 CO2 + 7 H2O
2 C7H14O2 +19 O2
21 O
14 CO2 + 14 H2O
(d) ? C2H4O2 + ? O2
? CO2 + ? H2O
C2H4O2 + ? O2
2 CO2 + ? H2O
2C
C2H4O2 + ? O2
2 CO2 + 2 H2O
4H
C2H4O2 + 2 O2
2 CO2 + 2 H2O
6O
Check your answers: (a) 6 C, 12 H & 18 O, (b) 5 C, 12 H & 16 O
(c) 14 C, 28 H, 42 O (d) 2 C, 4 H, 6 O
34. Define the problem: Balance the given equations.
Develop a plan: Follow the method described in the answers to Question 28.
Execute the plan:
(a) ? Mg + ? HNO3
? H2 + ? Mg(NO3)2
Select order: Mg, N, O, H
Mg + ? HNO3
? H2 + Mg(NO3)2
1 Mg
Mg + 2 HNO3
? H2 + Mg(NO3)2
2 N and 6 O
Mg + 2 HNO3
H2 + Mg(NO3)2
2H
(b) ? Al + ? Fe2O3
? Al2O3 + ? Fe
Select order: Fe, Al, O
? Al + Fe2O3
? Al2O3 + 2 Fe
2 Al + Fe2O3
Al2O3 + 2 Fe 2 Al and 3 O
2 Fe
(c) ? S + ? O2
? SO3
? S + 3 O2
2 SO3
6O
2 S + 3 O2
2 SO3
2S
(d) ? SO3 + ? H2O
SO3 + ? H2O
SO3 + H2O
? H2SO4
H2SO4
H2SO4
Select order: O, S
Select order: S, H, O
1S
2 H and 4 O
Chapter 4: Quantities of Reactants and Products
137
Check your answers: (a) 1 Mg, 2 H, 2 N & 6 O, (b) 2 Al, 2 Fe & 3 O
(c) 2 S & 6 O, (d) 1 S, 4 O & 2 H
The Mole and Chemical Reactions
35. Define the problem: Given a balanced chemical equation and the number of moles of a product formed,
determine the moles of reactant that was needed.
Develop a plan: Use the stoichiometry of the balanced equation as a conversion factor to convert the moles
of product to moles of reactant.
Execute the plan: The balanced equation says: 4 mol HCl are needed to make 1 mol Cl2.
12.5 mol Cl 2 ×
4 mol HCl
= 50.0 mol HCl
1 mol Cl 2
Check your answer: More HCl is needed than the Cl2 is formed. It makes sense that the moles of HCl is
greater.
36. Define the problem: Given a balanced chemical equation and the number of moles of a reactant, determine
the moles of another reactant needed.
Develop a plan: Use the stoichiometry of the balanced equation as a conversion factor to convert the moles
of the one reactant to moles of the other reactant.
Execute the plan: The balanced equation says: 1 mol CH4 reacts with 2 mol O2.
16.5 mol CH 4 ×
2 mol O2
= 33.0 mol O 2 needed
1 mol CH 4
Check your answer: More O2 is needed than the CH4. It makes sense that the moles of O2 is greater.
37. Define the problem: Given a balanced chemical equation and the number of moles of a reactant, determine
the moles and grams of another reactant needed and the grams of product produced.
Develop a plan: Use the stoichiometry of the balanced equation as a conversion factor to convert the moles
of the one reactant to moles of the other reactant. Use the molar mass to convert from moles to grams. Use
the stoichiometry of the equation to find the moles of product formed, then use the molar mass of the
product to find the grams.
Execute the plan: The balanced equation says: 2 mol NO react with 1 mol O2.
2.2 mol NO ×
1 mol O 2
= 1.1 mol O 2
2 mol NO
1.1 mol O2 ×
31.9988 g O2
= 35 g O2
1 mol O2
The balanced equation says: 2 mol NO produces 2 mol NO2.
2.2 mol NO×
2 mol NO2 46.0055 g NO 2
2
×
= 1.0 ×10 g NO2 produced
2 mol NO
1 mol NO2
138
Chapter 4: Quantities of Reactants and Products
Check your answers: Fewer moles of O2 are needed than moles of NO, so that answer makes sense. The
mass of NO used is 2.2 mol × (30.0061 g/mol) = 66 g. The sum of the reactant masses (66 g + 35 g) is equal to
2
the products mass (1.0 × 10 g), within known significant figures. These numbers make sense.
38. Define the problem: Given a balanced chemical equation and the number of moles of a reactant, determine
the moles and grams of another reactant needed and the grams of product produced.
Develop a plan: Use the stoichiometry of the balanced equation as a conversion factor to convert the moles
of the one reactant to moles of the other reactant. Use the molar mass to convert from moles to grams. Use
the stoichiometry of the equation to find the moles of product formed, then use the molar mass of the
product to find the grams.
Execute the plan: The balanced equation says: 4 mol Al react with 3 mol O2.
6.0 mol Al ×
4.5 mol O2 ×
3 mol O2
= 4.5 mol O2
4 mol Al
31.9988 g O 2
2
= 1.4× 10 g O 2
1 mol O 2
The balanced equation says: 4 mol Al produces 2 mol Al2O3.
6.0 mol Al ×
1 mol Al2 O3 101.9612 g Al 2O 3
2
×
= 3.1×10 g Al2O3 produced
2 mol Al
1 mol Al 2O 3
Check your answers: More moles of O2 are needed than moles of Al, so that answer makes sense. The
2
2
mass of Al used is 6.0 mol × (26.9815 g/mol) = 1.6 × 10 g. The sum of the reactant masses (1.6 × 10 g + 1.4
2
2
× 10 g) are equal to the products mass (3.1 × 10 g), within the uncertainty of the significant figures. These
numbers make sense.
39. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine
the grams of another reactant needed and the moles and grams of product produced.
Develop a plan: Use the molar mass of the first reactant to find the moles of that substance. Then use the
stoichiometry of the balanced equation as a conversion factor to convert the moles of the one reactant to
moles of the other reactant. Use the molar mass to convert from moles to grams. Use the stoichiometry of
the equation to find the moles of product formed, then use the molar mass of the product to find the grams.
Execute the plan: The balanced equation says: 1 mol Fe reacts with 1 mol Cl2.
10 .0 g Fe ×
0.179 mol Fe ×
1 mol Fe
= 0.179 mol Fe
55.845 g Fe
1 mol Cl 2 70.906 g Cl 2
×
= 12.7 g Cl 2
1 mol Fe
1 mol Cl 2
The balanced equation says: 1 mol Fe produces 1 mol FeCl2.
0.179 mol Fe×
1 mol FeCl2
= 0.179 mol FeCl2 expected
1 mol Fe
Chapter 4: Quantities of Reactants and Products
0.179 mol FeCl 2 ×
139
126.751 g FeCl 2
= 22.7 g FeCl 2 expected
1 mol FeCl 2
Check your answers: The sum of the masses of the reactants must add up to the total mass of the product.
10.0 g + 12.7 g = 22.7 g. This is the same product mass as that calculated above. These numbers make
sense.
40. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine
the grams of another reactant needed and the moles and grams of product produced.
Develop a plan: Use the molar mass of the first reactant to find the moles of that substance. Then use the
stoichiometry of the balanced equation as a conversion factor to convert the moles of the one reactant to
moles of the other reactant. Use the molar mass to convert from moles to grams. Use the stoichiometry of
the equation to find the moles of product formed, then use the molar mass of the product to find the grams.
Execute the plan:
(a) The balanced equation says: 2 mol Mn react with 3 mol F2.
5.12 g Mn ×
0.0932 mol Mn ×
1 mol Mn
= 0.0932 mol Mn
54.9380 g Mn
3 mol F2 37.9996 g F 2
×
= 5.31 g F2
2 mol Mn
1 mol F 2
(b) The balanced equation says: 2 mol Mn produces 2 mol MnF3.
0.0932 mol Mn ×
0.0932 mol MnF3 ×
1 mol F2
= 0.0932 mol MnF3 expected
1 mol Mn
111.9332 g MnF3
= 10.4 g MnF3 expected
1 mol MnF3
Check your answers: The sum of the masses of the reactants must add up to the total mass of the product.
5.12 g + 5.31 g = 10.43 g. This is the same mass (within three significant figures) as that calculated in part
(b). These numbers make sense.
41. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine
the moles of the reactant and the mass and moles of two products.
Develop a plan: Use the molar mass of the reactant to find the moles of that substance. Then use the
stoichiometry of the balanced equation as a conversion factor to convert the moles of reactant to moles of
each product formed, then use the molar mass of each product to find the grams.
Execute the plan: Molar mass of (NH4)2PtCl6 =
2 × [1 × (14.0067 g) + 4 × (1.0079 g)] + (195.078 g) + 6 × (35.453 g) = 443.873 g/mol
12.35 g ( NH 4 ) 2 PtCl 6 ×
1 mol (NH 4) 2 PtCl 6
−2
= 2.782 × 10 mol ( NH 4 ) 2 PtCl 6
443.873 g (NH 4) 2 PtCl 6
The balanced equation says: 3 mol (NH4)2PtCl6 react to form 3 mol Pt.
140
Chapter 4: Quantities of Reactants and Products
2.782 × 10
−2
mol (NH 4 ) 2 PtCl 6 ×
1 mol Pt
−2
= 2.782 × 10 mol Pt
1 mol ( NH 4) 2 PtCl 6
2.782 × 10−2 mol Pt ×
195 .078 g Pt
= 5.428 g Pt
1 mol Pt
The balanced equation says: 3 mol (NH4)2PtCl6 react to form 16 mol HCl.
2.782 × 10
−2
16 mol HCl
mol (NH 4 ) 2 PtCl 6 ×
= 0.1484 mol HCl
3 mol ( NH4 ) 2 PtCl 6
0.1484 mol HCl ×
36.461 mol HCl
= 5.410 g HCl
1 mol HCl
The complete table looks like this:
(NH4)2PtCl 6
Pt
HCl
12.35 g
5.428 g
5.410 g
0.02782 mol
0.02782 mol
0.1484 mol
Check your answers: The moles are smaller than the grams. This is appropriate. The moles of HCl are larger
than the moles of (NH4)2PtCl6 and Pt. These numbers make sense.
42. Define the problem: Given a balanced chemical equation and the number of grams of a product, determine
the moles of the product and the mass and moles of the two reactants.
Develop a plan: Use the molar mass of the product to find the moles of that substance. Then use the
stoichiometry of the balanced equation as a conversion factor to convert the moles of product to moles of
each reactant, then use the molar mass of each reactant to find the grams.
Execute the plan:
1 mol S 2Cl 2
103.5 g S 2 Cl 2 ×
= 0.7665 mol S 2 Cl 2
135.036 g S 2Cl 2
The balanced equation says: 4 mol S2Cl2 is produced from 1 mol S8.
1 mol S 8
0.7665 mol S 2Cl 2 ×
= 0.1916 mol S 8
4 mol S 2Cl 2
0.1916 mol S 8 ×
256.520 g S 8
= 49.15 g S 8
1 mol S8
The balanced equation says: 4 mol S2Cl2 is produced from 4 mol S8.
1 mol Cl 2
0.7665 mol S 2Cl 2 ×
= 0.7665 mol Cl 2
1 mol S 2 Cl 2
0.7665 mol Cl 2 ×
70.906 g Cl 2
= 54.35 g Cl 2
1 mol Cl 2
The complete table looks like this:
S8
Cl 2
S 2Cl 2
Chapter 4: Quantities of Reactants and Products
49.16 g
54.35 g
103.5 g
0.1916 mol
0.7665 mol
0.7665 mol
141
Check your answers: The sum of the masses of the reactants (49.16 g + 54.35 g = 103.51 g) adds up the mass
of the product given (103.5 g). The moles of S8 are smaller than the moles of Cl2 and S2Cl2, because of the
smaller stoichiometric coefficient. These numbers make sense.
43. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine
the moles of another reactant that are needed and the masses of the two products expected.
Develop a plan: Use the molar mass of the reactant to find the moles of that substance. Then use the
stoichiometry of the balanced equation as a conversion factor to convert the moles of one reactant to moles
of the other reactant. Use the stoichiometry to determine the moles of each of the products produced, then
use the molar mass of each product to find the grams.
Execute the plan:
(a) The balanced equation says: 1 mol TiCl4 reacts with 2 mol H2O.
14.0 g TiCl 4 ×
1 mol TiCl 4
= 0.0738 mol TiCl 4
189.688 g TiCl 4
2 mol H 2 O
0.0738 mol TiCl 4 ×
= 0.148 mol H 2O
1 mol TiCl
4
(b) The balanced equation says: 1 mol TiCl4 produces 1 mol TiO2.
1 mol TiO 2 79.866 g TiO 2
0.0738 mol TiCl 4 ×
×
= 5.89 g TiO 2
1 mol TiCl 4
1 mol TiO 2
The balanced equation says: 1 mol TiCl4 produces 4 mol HCl.
4 mol HCl 36.461 g HCl
0.0738 mol TiCl 4 ×
×
= 10.8 g HCl
1 mol TiCl 4
1 mol HCl
Check your answers: The sum of the masses of the reactants (14.0 g + 2.67 g = 16.7 g) adds up the mass of
the product (5.90 g + 10.8 g = 16.7 g). These numbers make sense.
44. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine
the moles of two other reactants needed and the mass of one product expected.
Develop a plan: Use the molar mass of the reactant to find the moles of that substance. Then use the
stoichiometry of the balanced equation as a conversion factor to convert the moles of one reactant to moles
of the other reactants. Use the stoichiometry to determine the moles of the product produced, then use the
molar mass of the product to find the grams.
Execute the plan:
(a) The balanced equation says: 2 mol SO2 reacts with 2 mol CaCO3.
150. g SO 2 ×
1 mol SO 2
1 mol CaCO 3
×
= 2.34 mol CaCO 3
64.064 g SO 2
1 mol SO 2
142
Chapter 4: Quantities of Reactants and Products
The balanced equation says: 2 mol SO2 reacts with 1 mol O2.
1 mol SO 2
1 mol O2
150. g SO 2 ×
×
= 1.17 mol O 2
64.064 g SO 2 2 mol SO 2
(b) The balanced equation says: 2 mol SO2 produces 2 mol CaSO4.
1 mol SO 2
1 mol CaSO 4 136.1406 g CaSO 4
150. g SO 2 ×
×
×
= 319 g CaSO 4
64.0638 g SO 2
1 mol SO 2
1 mol CaSO 4
Check your answers: The number of moles of O2 is half the number of moles of CaCO3. The calculated
moles are smaller than the calculated number of grams. These numbers make sense.
45. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine
the mass of one product formed.
Develop a plan: Convert from kilograms to grams using a standard metric conversion factor. Then use the
molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the balanced
equation to determine the moles of the product produced. Then use the molar mass of the product to find
the grams.
Execute the plan: The balanced equation says: 1 mol WO3 produces 1 mol W.
1.00 kg WO3 ×
1000 g WO3
1 mol WO3
1 mol W
183.84 g W
×
×
×
= 793 g W
1 kg WO3 231 .84 g WO3 1 mol WO3
1 mol W
Check your answer: It makes sense that the mass of W is smaller than the mass of WO3.
46. Define the problem: Given the formula of a compound and its mass, determine the mass of the elements
required to make it.
Develop a plan: Use the molar mass of the compound to find the moles of that substance. Then use the
stoichiometry of the balanced equation to determine the moles of each element needed. Then use the molar
masses of these elements to find the grams.
Execute the plan: The formula gives: 1 mol GaAs is produced from 1 mol Ga and 1 mol As.
1.45 g GaAs ×
1 mol GaAs
1 mol Ga
69 .723 g Ga
×
×
= 0.699 g Ga
144 .6446 g GaAs 1 mol GaAs
1 mol Ga
1.45 g GaAs ×
1 mol GaAs
1 mol As
74.9215 g As
×
×
= 0.751 g As
144 .6446 g GaAs 1 mol GaAs
1 mol As
Check your answers: The sum of the reactant masses (0.699 g + 0.751 g = 1.450 g) is the same as the total
mass of the compound. These numbers make sense.
47. Define the problem: Given a balanced chemical equation and the mass of the reactant, determine the mass
of each of the products produced.
Develop a plan: Use the molar mass of the compound to find the moles of that substance. Then use the
stoichiometry of the equation to determine the moles of the products produced. Then use the molar mass of
each product to find the grams.
Execute the plan: The balanced equation says: 2 mol NH4NO3 produces 2 mol N2.
Chapter 4: Quantities of Reactants and Products
143
1000 g NH 4NO 3
1 mol NH 4NO 3
2 mol N 2
×
×
1 kg NH 4 NO 3
80.0432 g NH 4NO 3 2 mol NH 4 NO 3
28.0134 g N 2
2
×
= 3.5 × 10 g N2
1 mol N 2
The balanced equation says: 2 mol NH4NO3 produces 4 mol H2O.
1.0 kg NH 4 NO 3 ×
1.0 kg NH 4 NO 3 ×
1000 g NH 4NO 3
1 mol NH 4NO 3
4 mol H 2O
×
×
1 kg NH 4 NO 3
80.0432 g NH 4NO 3 2 mol NH 4 NO 3
18.0152 g H 2O
2
×
= 4.5 × 10 g H2O
1 mol H2O
The balanced equation says: 2 mol NH4NO3 produces 1 mol O2.
1.0 kg NH 4 NO 3 ×
1000 g NH 4NO 3
1 mol NH 4NO 3
1 mol O 2
×
×
1 kg NH 4 NO 3
80.0432 g NH 4NO 3 2 mol NH 4 NO 3
31.9988 g O2
2
×
= 2.0 × 10 g O2
1 mol O2
Check your answers: The sum of the product masses
2
2
2
3
3.5 × 10 g + 4.5 × 10 g + 2.0 × 10 g = 1.00 × 10 g
is the same as the total mass of the reactant compound. These numbers make sense.
48. Define the problem: Given the products and reactants of a reaction and the mass of one reactant, balance
the chemical equation, determine the mass of the product produced, and determine the mass of the other
reactant.
Develop a plan: Balance the equation from the given formulas of the reactants and product. Use the molar
mass of the reactant to find the moles of that substance. Then use the stoichiometry of the equation to
determine the moles of the product produced and the moles of the other reactant required. Then use the
molar mass of each to find the grams.
Execute the plan: (a)
Balance O atoms, then Fe atoms.
4 Fe(s) + 3 O2(g)
2 Fe2O3(s)
(b) The balanced equation says: 4 mol Fe produces 2 mol Fe2O3.
5.58 g Fe ×
1 mol Fe
2 mol Fe 2O3 159.688 g Fe 2O 3
×
×
= 7.98 g Fe2 O3
55.845 g Fe
4 mol Fe
1 mol Fe 2O3
(c) The balanced equation says: 4 mol Fe reacts with 3 mol O2.
5.58 g Fe ×
1 mol Fe
3 mol O 2 31.9988 g O2
×
×
= 2.40 g O 2
55.845 g Fe 4 mol Fe
1mol O2
Check your answers: The sum of the reactant masses (5.58 g + 2.40 g = 7.98 g)
is the same as the total mass of the product compound. These numbers make sense.
49. Define the problem: Given a balanced chemical equation and the mass of the reactant, determine the masses
of all the products produced.
144
Chapter 4: Quantities of Reactants and Products
Develop a plan: Use the molar mass of the reactant to find the moles of that substance. Then use the
stoichiometry of the equation to determine the moles of the products produced. Then use the molar mass of
each product to find the grams.
Execute the plan:
1.0 g C 3H 5( NO3 ) 3 ×
1 mol C3H 5 (NO 3 ) 3
= 0.0044 mol C3H5 (NO 3) 3
227.0863 g C 3H5( NO 3) 3
The balanced equation says: 4 mol C3H5(NO3)3 produces 12 mol CO2.
0.0044 mol C3H5 (NO 3) 3 ×
12 mol CO 2
44.0095 g CO2
×
= 0.58 g CO2
4 mol C3H 5( NO3) 3
1 mol CO 2
The balanced equation says: 4 mol C3H5(NO3)3 produces 10 mol H2O.
10 mol H2O
18.0152 g H2O
0.0044 mol C3 H5 (NO 3 ) 3 ×
×
= 0.20 g H2O
4 mol C3H 5 (NO 3) 3
1 mol H2 O
The balanced equation says: 4 mol C3H5(NO3)3 produces 6 mol N2.
6 mol N 2
28.0134 g N2
0.0044 mol C3H5 (NO 3) 3 ×
×
= 0.19 g N2
4 mol C 3H 5( NO3 ) 3
1 mol N2
The balanced equation says: 4 mol C3H5(NO3)3 produces 1 mol O2.
1 mol O2
31.9988 g O2
0.0044 mol C3H5 (NO 3) 3 ×
×
= 0.035 g O2
4 mol C 3H5( NO3 ) 3
1 mol O2
Check your answers: The sum of the product masses (0.58 g + 0.20 g + 0.19 g + 0.035 g = 1.01 g) is the same
(to two significant figures) as the total mass of the reactant compound (1.0 g). These numbers make sense.
50. Define the problem: Given the reactants and products of a reaction and the mass of a reactant, balance the
chemical equation, determine the mass of another reactant, and determine the mass of one product
produced.
Develop a plan: Given the formulas of the reactants and products, balance the equation, selecting
appropriate order for the systematic balancing of all the atoms. Then use the molar mass of the reactant to
find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the
other reactant and product produced. Then use the molar mass of each to find the grams.
Execute the plan:
(a) ? CCl2F2 + ? Na 2C2O4
? C + ? CO2 + ? NaCl + ? NaF
Select order: Cl, F, Na, O, C
CCl2F2 + ? Na 2C2O4
? C + ? CO2 + 2 NaCl + ? NaF
2 Cl
CCl2F2 + ? Na 2C2O4
? C + ? CO2 + 2 NaCl + 2 NaF
2F
CCl2F2 + 2 Na 2C2O4
? C + ? CO2 + 2 NaCl + 2 NaF
4 Na
CCl2F2 + 2 Na 2C2O4
? C + 4 CO2 + 2 NaCl + 2 NaF
8O
Chapter 4: Quantities of Reactants and Products
CCl2F2 + 2 Na 2C2O4
C + 4 CO2 + 2 NaCl + 2 NaF
145
5C
(b) The balanced equation says: 1 mol CCl2F2 requires 2 mol Na 2C2O4.
1 mol CCl 2 F2
76.8 g CCl 2 F2 ×
= 0.635 mol CCl 2F2
120.914 g CCl 2F 2
0.635 mol CCl 2F2 ×
2 mol Na 2C 2O4 133.9986 g Na 2 C2O 4
×
= 170. g Na 2C2 O4
1 mol CCl 2 F2
1 mol Na 2C2 O4
The balanced equation says: 1 mol CCl2F2 produces 4 mol CO2.
0.635 mol CCl 2F2 ×
4 mol CO 2
44.0095 g CO2
×
= 112 g CO 2
1 mol CCl 2 F 2
1 mol CO2
Check your answers: (a) Check the number of atom of each type in the reactants and products: 5 C, 2 Cl, 2
F, 4 Na, 8 O. The equation is properly balanced. (b) The masses of Na 2C2O4 and CO2 are both larger than
the mass of CCl2F2. That makes sense, looking at the stoichiometry and the molar masses.
51. Define the problem: Given the reactants and products of a reaction and the mass of a reactant, balance the
chemical equation, determine the mass of another reactant, and determine the mass of one product
produced.
Develop a plan: Given the formulas of the reactants and products, balance the equation, selecting
appropriate order for the systematic balancing of all the atoms. Then use the molar mass of the reactant to
find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the
other reactant and product produced. Then use the molar mass of each to find the grams.
Execute the plan:
(a) ? NH4NO3
? N2O + ? H2O
Select order: N, H, O
NH4NO3
N2O + ? H2O
2N
NH4NO3
N2O + 2 H2O
4 H and 3 O
(b) The balanced equation says: 1 mol NH4NO3 requires 1 mol N2O.
10.0 g NH 4 NO 3 ×
1 mol NH 4 NO 3
= 0.125 mol NH 4 NO 3
80.0432 g NH 4 NO 3
1 mol N2O
44.0128 g N 2O
0.125 mol NH 4 NO 3 ×
×
= 5.50 g N 2O
1 mol NH 4 NO 3
1 mol N 2O
The balanced equation says: 1 mol NH4NO3 requires 2 mol H2O.
2 mol H 2O
18.0152 g H 2O
0.125 mol NH 4 NO 3 ×
×
= 4.50 g H 2O
1 mol NH 4NO 3
1 mol H 2O
Check your answers: (a) Check the number of atom of each type in the reactants and products: 2 N, 4 H,
3 O. The equation is properly balanced. (b) The sum of the masses of the products (5.50 g + 4.50 g = 10.00
g) is equal to the mass of the reactant compound (10.0 g). These numbers make sense.
146
Chapter 4: Quantities of Reactants and Products
52. Define the problem: Given a balanced chemical equation and the mass of a reactant in kilograms, determine
the mass of one of the products produced and the mass of another reactant required.
Develop a plan: Use metric conversion factors to convert kilograms to grams. Use the molar mass of the
reactant to find the moles of that substance. Then use the stoichiometry of the equation to determine the
moles of the product produced and the other reactant required. Then use molar masses to find the grams.
Execute the plan: 1.00 kg Fe 2O3 ×
1000 g Fe2 O3
1 mol Fe2 O3
×
= 6.26 mol Fe2 O3
1 kg Fe 2O3
159.688 g Fe2 O3
(a) The balanced equation says: 1 mol Fe2O3 produces 1 mol Fe.
6.26 mol Fe 2O3 ×
2 mol Fe
55.845 g Fe
×
= 699 g Fe
1 mol Fe 2O 3
1 mol Fe
(b) The balanced equation says: 1 mol Fe2O3 requires 3 mol CO.
3 mol CO
28.0101 g CO
6.26 mol Fe 2O3 ×
×
= 526 g CO
1 mol Fe 2O 3
1 mol CO
Check your answers: The mass of iron produced is less than the mass of iron(III) oxide it was produced
from. The mass of CO used is less than the mass of iron(III) oxide even with a larger stoichiometric
coefficient because CO contains lighter-weight atoms.
53. Define the problem: Given the reactants and products of a reaction and the desired mass of a product,
balance the chemical equation and determine the mass of the reactants needed to produce that product.
Develop a plan: Given the formulas of the reactants and products, balance the equation, selecting
appropriate order for the systematic balancing of all the atoms. Then use the molar mass of the product to
find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the
reactants needed to make that much product. Then use the molar mass of each to find the grams.
Execute the plan:
(a) ? K2PtCl4 + ? NH3
? K2PtCl4 + 2 NH3
? Pt(NH3)2Cl2 + ? KCl
Select order: N, H, Pt, K, Cl
Pt(NH3)2Cl2 + ? KCl
2 N and 6 H
K2PtCl4 + 2 NH3
Pt(NH3)2Cl2 + ? KCl
1 Pt
K2PtCl4 + 2 NH3
Pt(NH3)2Cl2 + 2 KCl
2 K and 4 Cl
(b) The balanced equation says: 1 mol Pt(NH3)2Cl2 requires 1 mol K2PtCl4.
1 mol Pt( NH 3) 2Cl 2
2.50 g Pt(NH 3 ) 2 Cl 2 ×
= 0.00833 mol Pt (NH 3 ) 2 Cl 2
300.045 g Pt(NH 3 ) 2 Cl 2
0.00833 mol Pt( NH 3) 2 Cl 2 ×
1 mol K 2 PtCl 4
415.0866 g K 2PtCl 4
×
= 3.46 g K2PtCl4
1 mol Pt(NH 3 ) 2Cl 2
1 mol K 2PtCl 4
The balanced equation says: 1 mol Pt(NH3)2Cl2 requires 2 mol NH3.
0.00833 mol Pt( NH 3) 2 Cl 2 ×
2 mol NH 3
17.0304 g NH 3
×
= 0.284 g NH3
1 mol Pt(NH 3 ) 2 Cl 2
1 mol NH 3
Chapter 4: Quantities of Reactants and Products
147
Check your answers: (a) Check the number of atom of each type in the reactants and products: 2 K, 1 Pt, 2
N, 6 H, and 4 Cl. The equation is properly balanced. (b) The mass of K2PtCl4 is larger than the mass of
Pt(NH3)2Cl2, and the mass of NH3 is less than the mass of Pt(NH3)2Cl2. These both make sense, looking at
the stoichiometric relationships and the molar masses.
Limiting Reactant
54. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
limiting reactant and the mass of the product produced.
Develop a plan: Here, we use a slight variation of what the text calls “the mole method.” We will calculate a
directly comparable quantity, the moles of the desired product. Use the molar mass of the reactants to find
the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of
the product produced in each case. Identify the limiting reactant from the reactant that produces the least
number of products. Then, use the moles of product produced from the limiting reactant and the molar mass
of the product to find the grams.
Execute the plan: Exactly 1 gram of each reactant is present initially.
The balanced equation says: 1 mol Na 2SO4 produces 1 mol BaSO4.
1 mol Na 2 SO 4
1 mol BaSO 4
1 g Na 2SO 4 ×
×
= 0.00704016 mol BaSO 4
142.042 g Na 2SO 4 1 mol Na 2SO 4
The balanced equation says: 1 mol BaCl2 produces 1 mol BaSO4.
1 mol BaCl 2
1 mol BaSO 4
1 g BaCl 2 ×
×
= 0.00480231 mol BaSO 4
208.233 g BaCl 2 1 mol BaCl 2
The number of BaSO4 moles produced from BaCl2 is smaller (0.00480231 mol < 0.00704016 mol), so BaCl2 is
the limiting reactant.
Find the mass of 0.00480231 mol BaSO4:
0.00480231 mol BaSO 4 ×
233.390 g BaSO 4
= 1.12081 g BaSO 4
1 mol BaSO 4
Check your answers: There are fewer moles of BaSO4 present than BaCl2, and the equation needs the same
amount of BaSO4 as BaCl2, so it makes sense that BaCl2 is the limiting reactant.
55. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
excess reactant and the mass left over when the product produced.
Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances. Then use
the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the
excess reactant from the reactant that produces the most number of products. The reactant that produces
the least moles of product is the limiting reactant. From the mass of the limiting reactant, determine the
moles of the other reactant needed for complete reaction. Convert that number to grams using the molar
mass. Then subtract the quantity used from the initial mass given to get the mass of excess reactant.
Execute the plan: The balanced equation says: 2 mol Al produces 1 mol Al2O3.
148
Chapter 4: Quantities of Reactants and Products
100 . g Al ×
1 mol Al
1 mol Al2 O3
×
= 1.85 mol Al2O 3
26 .9815 g Al
2 mol Al
The balanced equation says: 3 mol MnO produces 1 mol Al2 O3 .
200 . g MnO×
1 mol MnO
1 mol Al2O 3
×
= 0.940 mol Al 2 O3
70.9374 g MnO 3 mol MnO
The number of Al2 O3 moles produced from Al is larger (1.85 mol > 0.940 mol), so Al is the excess reactant
and MnO is the limiting reactant.
The balanced equation says: 3 mol MnO is produced from 2 mol Al.
200 . g MnO×
1 mol MnO
2 mol Al 26 .9815 g Al
×
×
= 50.7 g Al used
70.9374 g MnO 3 mol MnO
1 mol Al
100. g Al initial – 50.7 g Al used up = 49 g Al remains unreacted
Check your answers: There are fewer moles of MnO present than Al, and the equation needs more MnO
than Al, so it makes sense that MnO is the limiting reactant. In addition, the calculation of excess reactant
remaining proved that the initial mass of Al was larger than required to react with all of the MnO.
56. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
limiting reactant, the mass of the product produced, and the mass remaining of the excess reactant when the
reaction is complete.
Develop a plan: Here, we use a slight variation of what the text calls “the mole method.” We will calculate a
directly comparable quantity, the moles of product. (a) Use the molar mass of the reactants to find the moles
of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the
product produced in each case. Identify the limiting reactant from the reactant that produces the least
number of products. (b) Use the moles of product produced from the limiting reactant and the molar mass of
the product to find the grams. (c) From the limiting reactant quantity, determine the moles of the other
reactant needed for complete reaction. Convert that number to grams using the molar mass. Then subtract
the quantity used from the initial mass given to get the mass of excess reactant.
Execute the plan:
(a) The balanced equation says: 2 mol Al produces 2 mol AlCl3.
2.70 g Al ×
1 mol Al
2 mol AlCl 3
×
= 0.100 mol AlCl 3
26.9815 g Al
2 mol Al
The balanced equation says: 3 mol Cl2 produces 2 mol AlCl3.
1 mol Cl 2
2 mol AlCl 3
4.05 g Cl 2 ×
×
= 0.0381 mol AlCl 3
70.906 g Cl 2
3 mol Cl 2
The number of AlCl3 moles produced from Cl2 is smaller (0.0381 mol < 0.100 mol), so Cl2 is the limiting
reactant and Al is the excess reactant.
(b) Find the mass of 0.0381 mol AlCl3:
0.0381 mol AlCl 3 ×
133.341 g AlCl 3
= 5.08 g AlCl 3
1 mol AlCl 3
Chapter 4: Quantities of Reactants and Products
149
(c) The balanced equation says: 3 mol Cl2 react with 2 mol Al.
1 mol Cl 2
2 mol Al 26.9815 g Al
4.05 g Cl 2 ×
×
×
= 1.03 g Al
70.906 g Cl 2 3 mol Cl 2
1 mol Al
2.70 g Al initial – 1.03 g Al used up = 1.67 g Al remains unreacted
Check your answers: There are fewer moles of Cl2 present than Al, and the equation needs more Cl2 than
Al, so it makes sense that Cl2 is the limiting reactant. In addition, the calculation in (c) proved that the initial
mass of Al was larger than required to react with all of the Cl2. The sum of the masses of the reactants that
reacted (4.05 g + 1.03 g = 5.08 g) equals the mass of the product produced (5.08 g). These answers make
sense.
57. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
limiting reactant and the mass of the product produced.
Develop a plan: (a) Use the molar mass of the reactants to find the moles of the reactant substances. Then
use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify
the limiting reactant from the reactant that produces the least number of products. (b) Use the moles of
product produced from the limiting reactant and the molar mass of the product to find the grams.
Execute the plan:
(a) The balanced equation says: 2 mol H2 produces 2 mol H2O.
100. g H 2 ×
1 mol H2
2 mol H2O
×
= 49.6 mol H 2O
2.0158 g H2 2 mol H2
The balanced equation says: 1 mol O2 produces 2 mol H2O.
1 mol O2
2 mol H 2O
100. g O2 ×
×
= 6.25 mol H 2O
31.9988 g O2
1 mol O2
The number of H2O moles produced from O2 is smaller (6.25 mol < 49.6 mol), so O2 is the limiting
reactant and H2 is the excess reactant.
(b) Find the mass of 6.25 mol H2O:
6.25 mol H2O ×
18.0152 g H2O
= 113 g H 2O
1 mol H 2O
Check your answers: The reactants are present in the same mass quantities, but the molar mass of H2 is far
smaller than the molar mass of O2, so there are far fewer O2 molecules. Therefore, it makes sense that O2 is
the limiting reactant.
58. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
limiting reactant, the mass of the product produced, and the mass remaining of the excess reactant when the
reaction is complete.
Develop a plan: (a) Use the molar mass of the reactants to find the moles of the reactant substances. Then
use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify
the limiting reactant from the reactant that produces the least number of products. (b) From the limiting
reactant quantity determine the moles of the other reactant needed for complete reaction. Convert that
150
Chapter 4: Quantities of Reactants and Products
number to mass using the molar mass. Then subtract the quantity used from the initial mass given, to get
the mass of excess reactant. (c) Use the moles of product produced from the limiting reactant and the molar
mass of the product to find the grams.
Execute the plan: (a) The balanced equation says: 1 mol CO produces 1 mol CH3OH.
74 .5 g CO ×
1 mol CO
1 mol CH3OH
×
= 2.66 mol CH 3OH
28.0101 g CO
1 mol CO
The balanced equation says: 2 mol H2 produces 1 mol CH3OH.
1 mol H2
1 mol CH3OH
12.0 g H 2 ×
×
= 2.98 mol CH 3OH
2.0158 g H
2 mol H
2
2
The number of CH3OH moles produced from CO is smaller (2.66 mol < 2.98 mol), so CO is the limiting
reactant and H2 is the excess reactant.
(b) The balanced equation says: 2 mol H2 react with 1 mol CO.
74.5 g CO ×
1 mol CO
2 mol H 2 2.0158 g H2
×
×
= 10.7 g H 2
28.0101 g CO 1 mol CO
1 mol H 2
12.0 g H2 initial – 10.7 g H2 used up = 1.3 g H2 remains unreacted
(c) Find the mass of 2.66 mol CH3OH:
2.66 mol CH3OH ×
32.0417 g CH 3OH
= 85.2 g CH 3OH
1 mol CH 3OH
Check your answers: The calculation in (b) proved that the initial mass of H2 was larger than required to
react with all of the CO. The sum of the masses of the reactants that reacted (74.5 g + 10.7 g = 85.2 g) equal
the mass of the product produced (85.2 g). These answers make sense.
59. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
moles of reactants and products present when the reaction is finished.
Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances present
initially. Then use the stoichiometry of the equation to determine the moles of one of the products. Identify
the limiting reactant from the reactant that produces the least number of products. From the limiting reactant
quantity, and using the stoichiometry of the balanced equation, determine the moles of the other reactant
and products.
Execute the plan: The balanced equation says: 1 mol CH4 produces 1 mol CO2.
1 mol CH 4
1 mol CO 2
995 g CH 4 ×
×
= 62.0 mol CO 2
16.0423 g CH 4 1 mol CH 4
The balanced equation says: 2 mol H2O produces 1 mol CO2.
1 mol H 2 O
1 mol CO2
2510 g H 2O×
×
= 69.6 mol CO 2
18.0152 g H2 O 2 mol H 2O
Chapter 4: Quantities of Reactants and Products
151
The number of CO2 moles produced from CH4 is smaller (62.0 mol < 69.6 mol), so CH4 is the limiting reactant
and H2O is the excess reactant.
The balanced equation says: 1 mol CO2 is produced with 4 mol H2.
4 mol H 2
62.0 mol CO2 ×
= 248 mol H 2
1 mol CO2
The balanced equation says: 1 mol CH4 reacts with 2 mol H2O.
1 mol CH 4
2 mol H2 O
995 g CH 4 ×
×
= 124 mol H 2O reacted
16.0423 g CH 4 1 mol CH 4
1 mol H2 O
2510 g H 2O×
= 139 mol H 2O present initially
18.0152 g H2 O
139 mol H2O initial – 124 mol H2O reacted = 15 mol H2O left
Check your answers: The excess moles of H2O prove that the right limiting reactant was determined, since
there are no moles of CH4 left. The moles of products are stoichiometrically appropriate multiples of
62.0 moles, as they should be. These numbers look right.
60. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
moles of reactants and products present when the reaction is finished.
Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances present
initially. Then use the stoichiometry of the equation to determine the moles of one of the products. Identify
the limiting reactant from the reactant that produces the least number of products. From the limiting reactant
quantity, and using the stoichiometry of the balanced equation, determine the moles of the other reactant
and products.
Execute the plan: The balanced equation says: 1 mol CaO produces 1 mol H2O.
112 g CaO ×
1 mol CaO
1 mol H 2 O
×
= 2.00 mol H 2O
56 .077 g CaO 1 mol CaO
The balanced equation says: 2 mol NH4Cl produces 1 mol H2O.
1 mol NH 4Cl
1 mol H2 O
224 g NH 4Cl ×
×
= 2.09 mol H 2O
53.4913 g NH 4Cl 2 mol NH 4 Cl
The number of H2O moles produced from CaO is smaller (2.00 mol < 2.09 mol), so CaO is the limiting reactant
and NH4Cl is the excess reactant.
The balanced equation says: 1 mol CaO produces 2 mol NH3.
112 g CaO ×
1 mol CaO
2 mol NH 3
×
= 4.00 mol NH 3
56 .077 g CaO 1 mol CaO
The balanced equation says: 1 mol CaO produces 1 mol CaCl2.
152
Chapter 4: Quantities of Reactants and Products
112 g CaO ×
1 mol CaO
1 mol CaCl 2
×
= 2.00 mol CaCl 2
56 .077 g CaO
1 mol CaO
The balanced equation says: 1 mol CaO reacts with 2 mol NH4Cl.
112 g CaO ×
1 mol CaO
2 mol NH 4Cl
×
= 4.00 mol NH 4Cl reacted
56 .077 g CaO
1 mol CaO
1 mol NH 4Cl
224 g NH 4Cl ×
= 4.19 mol NH 4Cl present initially
53.4913 g NH 4Cl
4.19 mol NH4Cl initial – 4.00 mol NH4Cl reacted = 0.19 mol NH4Cl left
Check your answers: The excess moles of NH4Cl prove that the right limiting reactant was determined, since
there are zero moles of CaO left. The moles of products are stoichiometrically appropriate multiples of
2.00 moles, as they should be. These numbers look right.
61. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
maximum amount of product that can be produced.
Develop a plan: Use a metric conversion and the molar mass of the reactants to find the moles of the
reactant substances present initially. Then use the stoichiometry of the equation to determine the moles of
one of the products. Identify the limiting reactant from the reactant that produces the least number of
products. From the limiting reactant quantity, determine the moles of the product, then use the molar mass
of the product to get the grams and a metric conversion to get kilograms.
Execute the plan: The balanced equation says: 1 mol Fe2O3 produces 2 mol Fe.
2.00 kg Fe2 O3 ×
1000 g Fe 2O3
1 mol Fe 2O3
2 mol Fe
×
×
= 25.0 mol Fe
1 kg Fe2 O3
159.6882 g Fe 2O 3 1 mol Fe2 O3
The balanced equation says: 3 mol CO produces 2 mol Fe.
2.00 kg CO ×
1000 g CO
1 mol CO
2 mol Fe
×
×
= 47.6 mol Fe
1 kg CO
28.0101 g CO 3 mol CO
The number of Fe moles produced from Fe2O3 is smaller (25.0 mol < 47.6 mol), so Fe2O3 is the limiting
reactant and CO is the excess reactant.
Find the mass of the Fe from the limiting reactant:
25 .0 mol Fe ×
55 .845 g Fe
1 kg Fe
×
= 1.40 kg Fe
1 mol Fe
1000 g Fe
Check your answer: The reactants are present in the same mass quantities, but Fe2O3 has a larger molar
mass. So, it makes sense that it is the limiting reactant. The mass of iron should be smaller than the mass of
iron (III) oxide. These numbers look right.
62. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
maximum amount of product that can be produced.
Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances present
initially. Then use the stoichiometry of the equation to determine the moles of one of the products. Identify
Chapter 4: Quantities of Reactants and Products
153
the limiting reactant from the reactant that produces the least number of products. From the limiting reactant
quantity, determine the moles of the product, then use the molar mass of the product to get the grams.
Execute the plan: The balanced equation says: 2 mol C7H6O3 produces 2 mol C9H8O4.
100. g C7 H 6O 3 ×
1 mol C7 H 6O 3
2 mol C 9H 8 O4
×
= 0.724 mol C 9H 8 O4
138.1205 g C 7 H6O3 2 mol C 7 H6O3
The balanced equation says: 1 mol C4H6O3 produces 2 mol C9H8O4.
100. g C4 H6 O3 ×
1 mol C4 H6O3
2 mol C 9H 8O 4
×
= 1.96 mol C 9H 8 O4
102.0884 g C 4H 6 O3 1 mol C 4H 6O 3
The number of C9H8O4 moles produced from C7H6O3 is smaller (0.724 mol < 1.96 mol), so C7H6O3 is the
limiting reactant and C4H6O3 is the excess reactant.
Find the mass of the C9H8O4 from the moles of limiting reactant:
0.724 mol C 9H 8O 4 ×
180.1571 g C9 H 8O 4
= 130. g C 9H 8O 4
1 mol C 9H 8O 4
Check your answers: The reactants are present in the same mass quantities, but C7H6O3 has a larger molar
mass and needs two moles rather than just one mole to react. So, it makes sense that it is the limiting
reactant.
Percent Yield
63. Define the problem: Given the balanced chemical equation, the mass of the limiting reactant and the actual
yield, determine the theoretical yield and the percent yield.
Develop a plan: First, calculate the theoretical yield by determining the maximum mass of product that could
have been made from the given quantity of reactant: Take the mass of the limiting reactant and convert it to
moles. Then use stoichiometry to find the moles of product. Then convert to grams using molar mass.
Take the given actual yield and divide by the calculated theoretical yield and multiply by 100 % to get
percent yield.
Execute the plan: The limiting reactant is Fe2O3. From its mass, find the maximum grams of Fe that could be
made. The mole ratio comes from the balanced equation.
1.00 kg Fe 2O3 ×
1000 g Fe2 O3
1 mol Fe2 O3
2 mol Fe
55.845 g Fe
×
×
×
= 699 g Fe
1 kg Fe 2O3
159.688 g Fe2 O3 1 mol Fe 2O 3
1 mol Fe
The given mass of Fe is the actual yield.
654 g Fe actual
× 100 % = 93.5 % yield
699 g Fe theoretical
Check your answer: Close to the maximum quantity of iron was produced, so it makes sense that the
percent yield is over ninety percent.
64. Define the problem: Given the theoretical yield and the actual yield, determine the percent yield.
154
Chapter 4: Quantities of Reactants and Products
Develop a plan: Divide the actual yield by the theoretical yield and multiply by 100 % to get percent yield.
(Note that the balanced equation is given too, but you don’t need to use it to answer this question.)
Execute the plan:
100 g NH 3 actual
×100 % = 73.5 % yield
136 g NH 3 theoretical
Check your answer: About three-quarters of the maximum quantity of ammonia was produced, so it makes
sense that the percent yield is about 75 %. This number looks right.
65. Define the problem: Given the theoretical yield and the actual yield, determine the percent yield.
Develop a plan: Divide the actual yield by the theoretical yield and multiply by 100 % to get percent yield.
(Note that the balanced equation is given, too, but you don’t need to use it to answer this question.)
Execute the plan
36.7 g CaO actual
×100 % = 56.0 % yield
65.5 g CaO theoretical
Check your answer: A little more than half the maximum quantity of quicklime was produced, so it makes
sense that the percent yield is a little more than 50 %. This number looks right.
66. Define the problem: Given the balanced chemical equation, the mass of the limiting reactant and the actual
yield, determine the percent yield.
Develop a plan: First, we need to calculate the theoretical yield. We get this by determining the maximum
amount of product that could have been made from the given quantities of reactant: Take the mass of the
limiting reactant and convert it to moles. Then use stoichiometry to find the moles of product. Then
convert to grams using molar mass. Take the given actual yield and divide by the calculated theoretical
yield and multiply by 100 % to get percent yield.
Execute the plan: The limiting reactant is NaBH4. From its mass, find the maximum grams of B2H6 that could
be made. The mole ratio comes from the balanced equation.
1.203 g NaBH 4 ×
1 mol NaBH 4
1 mol B2 H 6 27.669 g B2 H6
×
×
= 0.4399 g B 2 H 6
37.832 g NaBH 4 2 mol NaBH 4
1 mol B 2H 6
The given mass of B2H6 is the actual yield.
0.295 g B2 H 6 actual
× 100 % = 67.1 % yield
0.4399 g B2 H 6 theoretical
Check your answer: About two-thirds the maximum quantity of diborane was produced, so it makes sense
that the percent yield is about 67 %. This number looks right.
67. Define the problem: Given the balanced chemical equation, the mass of the limiting reactant and the actual
yield, determine the percent yield.
Develop a plan: First, we need to calculate the theoretical yield. We get this by determining the maximum
amount of product that could have been made from the given quantities of reactant: Take the mass of the
limiting reactant and convert it to moles. Then use stoichiometry to find the moles of product. Then
convert to grams using molar mass. Take the actual yield and divide by the calculated theoretical yield and
multiply by 100 % to get percent yield.
Execute the plan: The limiting reactant is H2. From its mass, find the maximum grams of CH3OH that could
be made. The mole ratio comes from the balanced equation.
Chapter 4: Quantities of Reactants and Products
3
5.0× 10 g H 2 ×
155
1 mol H 2
1 mol CH 3OH 32.0417 g CH 3OH
4
×
×
= 4.0 ×10 g CH3OH
2.0158 g H 2
2 mol H 2
1 mol CH 3OH
The given mass of CH3OH is the actual yield.
3.5×10 3 g CH 3OH actual
4
4.0 ×10 g CH 3OH theoretical
× 100 % = 8.8 % yield
Check your answer: The actual mass produced is roughly a factor of ten less than the maximum quantity of
ethanol produced, so it makes sense that the percent yield is around 10 %. This number looks right.
68. Define the problem: Given the balanced chemical equation, the desired mass of the product, and the
percent yield, determine the amount of limiting reactant that must be used.
Develop a plan: Interpret the percent yield as the relationship between the actual grams and theoretical
grams. Use that relationship to find the theoretical yield mass of the product. Then use stoichiometry to
find the moles of limiting reactant. Then, using molar mass, convert to actual grams of reactant needed.
Execute the plan: The 51 % percent yield tells us the following:
To make 51 grams of S2Cl2, we need to have enough limiting reactant to make 1.19 grams of S2Cl2.
100. g S 2Cl 2
1.19 g S 2Cl 2 actual×
= 2.3 g S 2 Cl 2
51 g S 2 Cl 2 actual
Use this mass of product to determine what mass of limiting reactant to use.
2.3 g S 2Cl 2 ×
1 mol S 2Cl 2
3 mol SCl 2 102.971 g SCl 2
×
×
= 5.3 g SCl 2
135.036 g S 2Cl 2 1 mol S 2Cl 2
1 mol S 2Cl 2
Check your answer: The yield suggests that we need to try to make about twice as much, so the 2.3 grams
makes sense. The stoichiometry is 3:1, so it makes sense that a larger mass of SCl2 is needed than the mass
of S2Cl2 formed. These numbers look right.
69. Define the problem: Given the balanced chemical equation, the desired mass of the product, and the
percent yield, determine the amount of limiting reactant that must be used.
Develop a plan: Interpret the percent yield as the relationship between the actual grams and theoretical
grams. Use that relationship to find the theoretical yield mass of the product. Then use stoichiometry to
find the moles of limiting reactant. Then using molar mass, convert to actual grams of reactant needed.
Execute the plan: The 92 % percent yield tells us the following:
To make 92 grams of Si3N4, we need to have enough limiting reactant to make 100. grams of Si3N4.
1000 g Si 3 N4
100. g Si 3 N4
3
1.0 kg Si 3 N4 actual ×
×
= 1.1× 10 g Si 3N 4
1 kg Si 3 N4
92 g Si 3 N 4 actual
Use this mass of product to determine what mass of limiting reactant to use.
3
1.1× 10 g Si 3 N4 ×
1 mol Si 3N 4
3 mol Si
28.0855 g Si
×
×
= 660 g Si
140.2833 g Si 3 N 4 1 mol Si3 N 4
1 mol Si
156
Chapter 4: Quantities of Reactants and Products
Check your answer: The yield suggests that we don’t need to try to make a lot more, so the 1.1 kilograms
makes sense. The product mass is smaller because Si has a significantly smaller molar mass than Si3N4.
These numbers look right.
Empirical Formulas
70. Define the problem: Given the percent mass of elements in a compound, determine the empirical formula.
Develop a plan: Choose a convenient sample mass of Sx Oy , such as 100.0 g. Find the mass of S and O in
the sample, using the given mass percent. Use the molar mass of the elements to find their moles, then use a
whole-number mole ratio to determine the empirical formula.
Execute the plan: 100.0 g of .Sx Oy contains 40.0 g S and 60.0 g O.
40.0 g S ×
1 mol S
= 1.25 mol S
32 .065 g S
60 .0 g O ×
1 mol O
= 3.75 mol O
15 .9994 g O
Set up mole ratio and simplify by dividing by the smallest number of moles:
1.25 mol S : 3.75 mol O
1S:3O
Use the whole number ratio for the subscripts in the formula. The empirical formula is SO3 .
Check your answer: The percent by mass of oxygen in the compound is somewhat greater than the percent
by mass of sulfur. Since sulfur atoms weigh more than oxygen atoms, it makes sense that there are several O
atom per S atom present in this formula. SO3 is a common oxide of sulfur.
71. Define the problem: Given the percent mass of elements in a compound, determine the empirical formula.
Develop a plan: Choose a convenient sample mass of Kx Cry Oz, such as 100.00 g. Find the masses of K, Cr
and O in the sample, using the given mass percent. Use the molar mass of the elements to find their mo les,
then use a whole-number mole ratio to determine the empirical formula.
Execute the plan: 100.0 g of .Kx Cry Oz contains 26.57 g K, 35.36 g Cr, and 38.07 g O.
26 .57 g K ×
1 mol K
= 0.6796 mol K
39.0983 g K
38 .07 g O ×
35 .36 g Cr ×
1 mol Cr
= 0.6801 mol Cr
51 .9961 g Cr
1 mol O
= 2.379 mol O
15 .9994 g O
Set up mole ratio and simplify by dividing by the smallest number of moles:
0.6796 mol K : 0.6801 mol Cr : 2.379 mol O
1 K : 1 Cr : 3.5 O
Multiply by 2 to get whole numbers:
2 K : 2 Cr : 7 O
Use the whole number ratio for the subscripts in the formula. The empirical formula is K2 Cr2 O7 .
Chapter 4: Quantities of Reactants and Products
157
Check your answer: The percent by mass of oxygen in the compound is somewhat greater than the percent
by mass of potassium or chromium. Since K and Cr atoms weigh more than oxygen atoms, it makes sense
that there are several O atom per K and Cr atom present in this formula. This formula coincides with
potassium dichromate, a common ionic compound.
72. Define the problem: Given the mass of a compound, the identity of the elements in the compound, and the
identity and masses of all the products produced, determine the empirical formula.
Develop a plan: Use the molar mass of the products to find their moles and use the stoichiometry of their
formulas to determine the moles of the elements in the compound. Then use a whole-number mo le ratio to
determine the empirical formula.
Execute the plan:
The compound is a hydrocarbon and contains only C and H: CiHj. Its combustion produced H2O and CO2.
Use molar mass and formula stoichiometry to determine the moles of C and H.
1.481 g CO 2 ×
1 mol CO2
1 mol C
×
= 0.03365 mol C
44.0095 g CO2 1 mol CO 2
0.303 g H2 O×
1 mol H 2 O
2 mol H
×
= 0.0336 mol H
18.0152 g H 2 O 1 mol H 2O
Set up mole ratio and simplify by dividing by the smallest number of moles:
0.03365 mol C : 0.0336 mol H
1C:1H
Use the whole number ratio for the subscripts in the formula. The empirical formula is CH.
Check your answer: These are not necessary calculations, but we can calculate the masses of each element,
C and H.
0.03365 mol C ×
12.0107 g C
= 0.4041 g C
1 mol C
0.0336 mol H ×
1.0079 g H
= 0.0339 g H
1 mol H
The sum of these masses (0.4041 g + 0.0339 g = 0.4380 g) adds up to the mass of the original compound
(0.438 g). This answer makes sense.
73. Define the problem: Given the mass of a compound, the identity of the elements in the compound, and the
identity and masses of all the products produced, determine the empirical formula.
Develop a plan: Use the molar mass of the products to find their moles and use the stoichiometry of their
formulas to determine the moles of the elements in the compound. Then use a whole-number mole ratio to
determine the empirical formula.
Execute the plan: The compound is a hydrocarbon and contains only C and H: CiHj. Its combustion
produced H2O and CO2. Use molar mass and formula stoichiometry to determine the moles of C and H.
1 mol CO 2
1 mol C
−3
0.379 g CO2 ×
×
= 8.61 × 10 mol C
44.0095 g CO 2 1 mol CO 2
0.1035 g H 2 O×
1 mol H2 O
2 mol H
−2
×
= 1.149 × 10 mol H
18.0152 g H2 O 1 mol H2O
158
Chapter 4: Quantities of Reactants and Products
Set up mole ratio and simplify by dividing by the smallest number of moles:
–3
–2
8.61 × 10 mol C : 1.149 × 10 mol H
1 C : 1.33 H
Multiply by 3, to get a whole number ratio:
3C:4H
Use the whole number ratio for the subscripts in the formula. The empirical formula is C3H4.
Check your answer: These are not necessary calculations, but we can calculate the masses of each element,
C and H.
8.61 ×10−3 mol C×
0.01149 mol H ×
12.0107 g C
= 0.103 g C
1 mol C
1.0079 g H
= 0.01158 g H
1 mol H
The sum of these masses (0.103 g + 0.01158 g = 0.115 g) add up to the mass of the original compound (0.115
g). This answer make sense.
74. Define the problem: Given the mass of a compound, the identity of the elements in the compound, and the
identity and masses of all the products produced, determine the empirical formula.
Develop a plan: Combustion uses oxygen. When the compound also contains oxygen, determine the
amount of oxygen after the other elements. Use the molar mass of the products to find their moles and use
the stoichiometry of their formulas to determine the moles of the elements that are not oxygen in the
compound. Find the masses of those elements, and subtract them from the total mass of the compound to
get the mass of oxygen in the compound. Then use the molar mass to calculate the moles of oxygen in the
compound. Then use a whole-number mole ratio to determine the empirical formula.
Execute the plan: The compound contains C, H, and O: CiHjOk. Its combustion produced H2O and CO2.
Use molar mass and formula stoichiometry to determine the moles of C and H. Use the whole number ratio
for the subscripts in the formula.
0.421 g CO2 ×
1 mol CO 2
1 mol C
−3
×
= 9.56 × 10 mol C
44.0095 g CO 2 1 mol CO2
1 mol H 2O
2 mol H
−2
0.172 g H 2O ×
×
= 1.91× 10 mol H
18.0152 g H 2O 1 mol H 2 O
Calculate the masses of C and H.
9.56 × 10−3 mol C ×
12.0107 g C
= 0.115 g C
1 mol C
0.0191 mol H ×
1.0079 g H
= 0.0192 g H
1 mol H
Calculate the masses of O by subtracting the masses of C and H from the given total compound mass.
0.236 g CiHjOk – 0.115 g C – 0.0192 g H = 0.102 g O
Calculate the moles of O.
0.102 g H 2O ×
1 mol O
= 6.37 × 10 −3 mol O
15 .9994 g O
Chapter 4: Quantities of Reactants and Products
159
Set up mole ratio and simplify by dividing by the smallest number of moles:
–3
–2
–3
9.56 × 10 mol C : 1.91 × 10 mol H : 6.37 × 10 mol O
1.5 C : 3 H : 1 O
Multiply by 2, to get a whole number ratio:
3 C : 6 H: 2 O
Use the whole number ratio for the subscripts in the formula. The empirical formula is C3H6O2.
Check your answer: The mole ratio came out very close to whole number values. This answer makes sense.
75. Define the problem: Given the mass of a compound, the identity of the elements in the compound, and the
identity and masses of all the products produced, determine the empirical formula.
Develop a plan: Combustion uses oxygen. When the compound also contains oxygen, determine the
amount of oxygen after the other elements. Use the molar mass of the products to find their moles and use
the stoichiometry of their formulas to determine the moles of the elements that are not oxygen in the
compound. Find the masses of those elements, and subtract them from the total mass of the compound to
get the mass of oxygen in the compound. Then calculate the moles of oxygen in the compound. Then use a
whole-number mole ratio to determine the empirical formula.
Execute the plan: The compound contains C, H, and O: CiHjOk. Its combustion produced H2O and CO2.
Use molar mass and formula stoichiometry to determine the moles of C and H. Use the whole number ratio
for the subscripts in the formula.
0.257 g CO 2 ×
1 mol CO 2
1 mol C
−3
×
= 5.84 × 10 mol C
44.0095 g CO 2 1 mol CO2
0.0350 g H2O ×
1 mol H 2O
2 mol H
−3
×
= 3.89 × 10 mol H
18.0152 g H 2O 1 mol H 2 O
Calculate the masses of C and H.
5.84 × 10−3 mol C ×
12 .0107 g C
= 0.0701 g C
1 mol C
3.89 × 10−3 mol H ×
1.0079 g H
=0.00392 g H
1 mol H
Calculate the masses of O by subtracting the masses of C and H from the given total compound mass.
0.105 g CiHjOk – 0.0701 g C – 0.00392 g H = 0.031 g O
Calculate the moles of O.
0.031 g H 2 O ×
1 mol O
= 1.9 ×10−3 mol O
15 .9994 g O
Set up mole ratio and simplify by dividing by the smallest number of moles:
–3
–3
–3
5.84 × 10 mol C : 3.89 × 10 mol H : 1.9 × 10 mol O
3.0 C : 2.0 H : 1.0 O
Use the whole number ratio for the subscripts in the formula. The empirical formula is C3H2O.
160
Chapter 4: Quantities of Reactants and Products
Check your answer: The mole ratio came out very close to whole number values. This answer make sense.
General Questions
76. Define the problem: Given the unbalanced chemical equation and the mass of the limiting reactant,
determine the mass of a product that can be isolated.
Develop a plan: Balance the equation. Use the molar mass of the reactant to find moles of reactant, then
use the stoichiometry of the equation to determine moles of product, then use the molar mass of the product
to get the mass of the product.
Execute the plan: Select a balance order: H, O, N, Cu
2 NH3 + 3 CuO
N2 + 3 Cu + 3 H2O
Check the balance: 2 N, 6 H, 3 Cu, 3 O
26.3 g NH 3 ×
1 mol NH 3
1 mol N2 28.0134 g N 2
×
×
= 21.6 g N 2
17.0304 g NH 3 2 mol NH 3
1 mol N2
Check your answer: It makes sense that the initial mass of NH3 is nearly the same as the mass of N2 created,
since the NH3 molar mass is about half that of N2 and the stoichiometry is 2:1.
77. Define the problem: Given the mass of the only reactant, the formula of the reactant with one element
unknown, a balanced equation showing its decomposition, and the mass and identity of one of the
products, determine the identity of the unknown element from a given list.
Develop a plan: Use the molar mass of the product to find moles of product, then use the stoichiometry of
the equation to determine moles of reactant. Then divide the grams of reactant by the calculated moles of
reactant to determine the molar mass of the reactant. Subtract the molar masses of the known elements in
the compound to determine the molar mass of the unknown element. Compare the molar masses of the
elements on the list and find the one that most closely matches it.
Execute the plan: Use mass of CO2 to find moles of MCO3
1 mol CO 2
1 mol MCO 3
−3
0.376 g CO2 ×
×
= 8.54 ×10 mol MCO 3
44.0095 g CO 2 1 mol CO 2
Divide given mass by moles:
1.056 g MCO 3
g
= 124
= molar mass of MCO3
−3
mol
8.54 × 10 mol MCO 3
Subtract the molar mass of C and three times the molar mass of O from this molar mass to find the molar mass
of M:
124 g MCO 3 12.0107 g C 3× 15.9994 g O
–
–
= 64 g M
mol MCO 3
mol MCO 3
mol MCO 3
The element in the given list with the closest molar mass to 64 g/mol is (b) Cu.
Check your answer: The molar mass of Cu (63.5 g/mol) is 64. None of the others in the list (Ni at 58.7 g/mol,
Zn at 65.4 g/mol, or Ba at 137.2 g/mol) are this close, though within ±1 g/mol Zinc almost qualifies. If we
Chapter 4: Quantities of Reactants and Products
161
carry more decimal places than strictly allowed by the rules of significant figures, the molar mass of MCO3 is
123.65 g/mol and the molar mass of M is 63.6 g/mol. This might make us feel better selecting Cu, although
these additional significant figures are actually unknown with the limited data provided.
78. Define the problem: Given the balanced chemical equation and the masses of both reactants, determine the
maximum yield of a product in grams.
Develop a plan: Find the limiting reactant by determining moles of a product made from each reactant. Then
using the molar mass of the product, determine the mass from the moles produced by the limiting reactant.
Execute the plan:
365 g UO3 ×
1 mol UO3
6 mol UF4
×
= 1.28 mol UF4
286.0271 g UO3 6 mol UO3
1 mol BrF3
6 mol UF4
365 g BrF3 ×
×
= 2.00 mol UF4
136.8992 g BrF3 8 mol BrF3
UO3 is the limiting reactant, since fewer moles of UF4 are produced from the UO3 reactant (1.28 mol) than
produced from the BrF3 reactant (2.00 mol). Therefore, use 1.28 mol UF4 and the molar mass to determine the
grams of UF4 produced:
1.28 mol UF4 ×
314.0225 g UF 4
= 401 g UF4
1 mol UF 4
Check your answer: It makes sense that UO3 is the limiting reactant, because the reactants are present in
equal masses, and UO3 has a larger molar mass than BrF3. Since the molar mass of the product is somewhat
larger than the molar mass of the limiting reactant and their stoichiometry is 1:1, it makes sense that slightly
larger mass of product is formed in this reaction.
79. Define the problem: Given the balanced chemical equation and the mass of one reactant and the moles of
the other reactant, determine the maximum theoretical mass of a product.
Develop a plan: Find the limiting reactant by determining moles of a product made from each reactant. Then
using the molar mass of the product, determine the mass from the moles produced by the limiting reactant.
Execute the plan:
1 mol ( NH 4 ) 2 PtCl 4
1 mol Pt(NH 3 ) 2 Cl 2
15.5 g (NH 4 ) 2 PtCl 4 ×
×
= 0.0416 mol Pt(NH3)2Cl2
372.967 g ( NH 4) 2 PtCl 4 1 mol (NH 4 ) 2 PtCl 4
0.15 mol NH 3 ×
1 mol Pt( NH 4 ) 2 Cl 2
= 0.075 mol Pt( NH 4) 2Cl 2
2 mol NH 3
(NH4)2PtCl4 is the limiting reactant, since fewer moles of Pt(NH3)2Cl2 are produced from the (NH4)2PtCl4
reactant (0.0416 mol) than produced from the NH3 reactant (0.075 mol). Therefore, use
0.0416 mol Pt(NH3)2Cl2 and the molar mass to determine the grams of Pt(NH3)2Cl2 produced:
0.0416 mol Pt (NH 3 ) 2 Cl 2 ×
300.0448 g Pt(NH 3) 2 Cl 2
= 12.5 g Pt (NH 3 ) 2 Cl 2
1 mol Pt (NH 3 ) 2 Cl 2
162
Chapter 4: Quantities of Reactants and Products
Check your answer: It is satisfying that (NH4)2PtCl4 is the limiting reactant, because that reactant contains
the expensive platinum metal and the experimenter would want to make sure all of that got used up. Since
the molar mass of the product is somewhat smaller than the molar mass of the limiting reactant and their
stoichiometry is 1:1, it makes sense that a slightly smaller mass of product is formed in this reaction.
80. Define the problem: Given the balanced chemical equation, the moles of one reactant, and the mass of the
other reactant, determine the maximum yield of a product in grams.
Develop a plan: Find the limiting reactant by determining moles of a product made from each reactant. Then
using the molar mass of the product, determine the mass from the moles produced by the limiting reactant.
Execute the plan:
2.19 × 10
−2
1.55 g NaBH 4 ×
mol H 2SO 4 ×
1 mol B2 H6
= 0.0219 mol B 2H 6
1 mol H2SO 4
1 mol NaBH 4
1 mol B 2H 6
×
= 0.0205 mol B 2H 6
37.8324 g NaBH 4 2 mol NaBH 4
NaBH4 is the limiting reactant, since fewer moles of B2H6 are produced from the NaBH4 reactant (0.0219 mol)
than produced from the H2SO4 reactant (0.0205 mol). Therefore, use 0.0219 mol B2H6 and the molar mass to
determine the grams of B2H6 produced:
0.0205 mol B2 H 6 ×
27.6694 g B2 H6
= 0.567 g B 2H 6
1 mol B2 H6
Check your answer: It is satisfying that NaBH4 is the limiting reactant, because that reactant contains the
boron element and the experimenter would want to make sure all of that got converted to product. Since the
molar mass of the product is somewhat smaller than the molar mass of the limiting reactant and their
stoichiometry is 2:1, it makes sense that slightly smaller mass of product is formed in this reaction.
81. Define the problem: Given the mass of a compound, the identity of the elements in the compound, and the
identity and masses of all the products produced, determine the empirical formula.
Develop a plan: Use the molar mass of the products to find their moles and use the stoichiometry of their
formulas to determine the moles of the elements in the compound. Then use a whole-number mole ratio to
determine the empirical formula.
Execute the plan:
The compound is SixOy. Its combustion produced SiO2 and H2O. Use molar mass and formula
stoichiometry to determine the moles of Si and H.
1 mol SiO 2
1 mol Si
11.64 g SiO2 ×
×
= 0.1937 mol Si
60.0843 g SiO 2 1 mol SiO 2
6.980 g H 2O ×
1 mol H 2O
2 mol H
×
= 0.7749 mol H
18.0152 g H 2O 1 mol H 2 O
Set up mole ratio and simplify by dividing by the smallest number of moles:
0.1937 mol Si : 0.7749 mol H
1 Si : 4.000 H
Chapter 4: Quantities of Reactants and Products
163
Use the whole number ratio for the subscripts in the formula. The empirical formula is SiH4.
Check your answer: These are not necessary calculations, but we can calculate the masses of each element,
Si and H.
0.1937 mol Si ×
28.0855 g Si
= 5.440 g Si
1 mol Si
0.7749 mol H ×
1.0079 g H
= 0.7810 g H
1 mol H
The sum of these masses (5.440 g + 0.7810 g = 6.221 g) add up to the mass of the original compound (6.22 g),
to the given significant figures. This answer makes sense.
82. Define the problem: Given the mass of a compound, the identity of the elements in the compound, the mass
of one of the products produced, and the identity of all products produced, determine the empirical formula.
Develop a plan: Use the molar mass of the products to find their moles and use the stoichiometry of their
formulas to determine the moles of the elements in the compound. Then use a whole-number mole ratio to
determine the empirical formula.
Execute the plan:
The compound is BxHy. Its combustion produced B2O3 and H2O. Use molar mass and formula
stoichiometry to determine the moles of B.
1 mol B2O3
2 mol B
0.422 g B2O3 ×
×
= 0.0121 mol B
69.6202 g B2O3 1 mol B 2O 3
Find the mass of B in the compound:
0.0121 mol B ×
10.811 g B
= 0.131 g B
1 mol B
Find the mass of H in the compound, by subtracting the mass of B from the compound sample mass:
0.148 g BxHy – 0.131 g B = 0.017 g H
Now, determine moles of H:
0.017 g H ×
1 mol H
= 0.017 mol H
1.0079 g H
Set up mole ratio and simplify by dividing by the smallest number of moles:
0.0121 mol B : 0.017 mol H
1 B : 1.4 H
If we multiply by 5, we get a whole-number mole ratio:
5B:7H
Use this whole number ratio, the empirical formula is B5H7.
However, the significant figures in the calculation are small, so the quantity of H has an uncertainty of ± 0.1.
That means it could also really be a ratio of 1 : 1.5 or 1 : 1.33, and the multipliers would be 2 or 3, respectively
to give whole-number mole ratios of:
2B:3H
They have empirical formulas of B2H3 and B3H4.
or
3B:4H
164
Chapter 4: Quantities of Reactants and Products
Check your answer: The first sentence of the question gives us the impression that many boron hydride
compounds exist. The limited significant figures here do not allow us to rule out any of these three empirical
formulas: B5H7, B2H3, or B3H4.
83. Define the problem: Given equal moles of all the reactants, determine the limiting reactant. Given equal
masses of all the reactants, determine the limiting reaction.
Develop a plan: When the same moles of every reactant is present, the reactant with the largest
stoichiometric coefficient is going to be the limiting reactant. In the second part, use the molar mass of each
reactant to find their moles, then use the stoichiometry of the reaction to determine the moles of a product
formed from each reactant. The reactant that produces the least amount of product is the limiting reactant.
Execute the plan:
First Part: 5 moles of each reactant is present. The stoichiometric coefficients are: 4 for KOH, 2 for MnO2, 1
for O2 and 1 for Cl2. The reactant with the largest stoichiometric coefficient is KOH. So, KOH is the limiting
reactant.
Second Part: 5 grams of each reactant is present. Determine the moles of KCl they each form:
5 g KOH ×
5 g MnO2 ×
1 mol KOH
2 mol KCl
×
= 0.04 mol KCl
56 .1056 g KOH 4 mol KOH
1 mol MnO2
2 mol KCl
×
= 0.06 mol KCl
86.9368 g MnO2 2 mol MnO2
1 mol O2
2 mol KCl
5 g O2 ×
×
= 0.3 mol KCl
31.9988 g O2 1 mol O2
1 mol Cl 2
2 mol KCl
5 g Cl 2 ×
×
= 0.1 mol KCl
70.906 g Cl 2 1 mol Cl 2
Only 0.04 mol of HCl is produced from KOH, so the KOH is the limiting reactant here, too.
Check your answers: The first part is easy. The reaction says that more moles of KOH are needed that any
other reactant, so when equal moles of reactants are present, it runs out first. The differences in molar mass
are insufficient to keep the large stoichiometric coefficient for KOH from making it the limiting reactant when
equal masses of the reactants are present. These answers make sense.
84. Define the problem: Given equal moles of all the reactants, determine the limiting reactant. Given equal
masses of all the reactants, determine the limiting reaction.
Develop a plan: When the same moles of every reactant is present, the reactant with the largest
stoichiometric coefficient is going to be the limiting reactant. In the second part, use the molar mass of each
reactant to find their moles, then use the stoichiometry of the reaction to determine the moles of a product
formed from each reactant. The reactant that produces the least amount of product is the limiting reactant.
Execute the plan:
(a) 10 moles of each reactant is present. The stoichiometric coefficients are: 4 for NaCl, 2 for SO2, 2 for H2O
and 1 for O2. The reactant with the largest stoichiometric coefficient is NaCl. So, NaCl is the limiting
reactant.
Chapter 4: Quantities of Reactants and Products
165
(b) 100 grams of each reactant is present. (note: 100 has just one significant figure) Determine the moles
of KCl they each form:
1 mol NaCl
4 mol HCl
×
= 1.71 mol HCl ≈ 2 mol HCl
58.4428 g NaCl 4 mol NaCl
100 g NaCl ×
100 g SO 2 ×
100 g H 2O ×
1 mol SO 2
4 mol HCl
×
= 3.12 mol HCl ≈ 3 mol HCl
64.0683 g SO 2 2 mol SO 2
1 mol H2O
4 mol HCl
×
= 11.1 mol HCl ≈ 10 mol HCl
18.0152 g H2O 2 mol H 2O
1 mol O2
4 mol HCl
100 g O 2 ×
×
= 12.5 mol HCl ≈ 10 mol HCl
31.9988 g O2 1 mol O2
Only 2 mol of HCl is produced from NaCl, so the NaCl is the limiting reactant here, too.
Check your answers: (a) The reaction says that more moles of NaCl are needed that any other reactant, so
when equal moles of reactants are present, it runs out first. (b) The differences in molar mass are insufficient
to keep the large stoichiometric coefficient for NaCl from making it the limiting reactant when equal masses
of the reactants are present. These answers make sense.
Applying Concepts
85. Two butane molecules react with 13 diatomic oxygen molecules to produce eight carbon dioxide molecules
and ten water molecules.
Two moles of gaseous butane molecules react with 13 moles of gaseous diatomic oxygen molecules to
produce eight moles of gaseous carbon dioxide molecules and ten moles of liquid water molecules.
86. One P4 molecule reacts with six diatomic chlorine molecules to produce four PCl3 molecules.
One mole of solid P4 molecules reacts with six moles of gaseous diatomic chlorine molecules to produce four
moles of liquid PCl3 molecules.
87. Balance the equation for this reaction: 4 A 2 + AB3
Reactant number of A atoms = 4 × 2 + 1 = 9
Reactant number of B atoms = 3
3
Product number of A atoms = 9 = 3 × 3
Product number of A atoms = 3 = 3 × 1
So, the product molecule has 3 A atoms and 1 B atom. Its formula is A 3B.
4 A 2 + AB3
88. Balance the equation for this reaction: 3 A 2B3 + B3
Reactant number of A atoms = 3 × 2 = 6
3 A 3B
6
Product number of A atoms = 6 = 6 × 1
Reactant number of B atoms = 3 × 3 + 3 = 12
Product number of A atoms = 12 = 6 × 2
So, the product molecule has 1 A atom and 2 B atoms. Its formula is AB2.
3 A 2B3 + B3
6 AB2
Chapter 4: Quantities of Reactants and Products
166
89. Define the problem: Given the moles of all the reactants, determine the limiting reactant and the ions
present at the end or the reaction.
Develop a plan: Use the stoichiometry of the reaction to determine the moles of a product formed from each
reactant. The reactant that produces the least amount of product is the limiting reactant. The excess
reactant and the products are present at the end of the reaction. From that information, determine what ions
are in the solution after the reaction is complete.
1.5 mol Cu ×
Execute the plan:
1 mol Cu( NO 3) 2
= 1.5 mol Cu(NO 3) 2
1 mol Cu
4.0 mol AgNO 3 ×
1 mol Cu(NO 3 ) 2
= 2.0 mol Cu(NO 3 ) 2
2 mol AgNO 3
Only 1.5 mol Cu(NO3)2 can be formed, so the limiting reactant is Cu and the excess reactant is AgNO3.
+
2+
–
That means the solution contains: Ag ions, Cu ions, and NO3 ions after the reaction is over. (The
question only asked what ions are present not how many.)
If we want to know how many, we need to determine the moles of AgNO3 actually reacted:
2 mol AgNO 3
1.5 mol Cu ×
= 3.0 mol AgNO 3 reacted
1 mol Cu
The moles of AgNO3 left = 4.0 mol AgNO3 initial – 3.0 AgNO3 mol reacted = 1.0 mol AgNO3 left
+
Quantitatively, the solution has 1.0 mol Ag ions, 1.5 mol Cu
2+
ions, and (1.0 mol + 2 × 1.5 mol =) 4.0 mol
–
NO3 ions.
Check your answers: The excess reactant is an ionic compound and one of the products is an ionic
compound, so some of the ions present at the beginning are still present when the Cu runs out. The
+
quantities of Ag ions have dropped, since some of the silver atoms are now part of the Ag solid product.
–
The quantity of NO3 ions doesn’t change, since neither N nor O are part of the solid product formed.
These answers make sense.
90. Define the problem: Given the number of molecules of all the reactants, determine what molecules are
present after a reaction is complete and determine the limiting reactant.
Develop a plan: Use the stoichiometry of the reaction to determine the molecules of the product formed
from each reactant. The reactant that produces the least amount of product is the limiting reactant. The
excess reactant and the products are present at the end of the reaction. From that information, determine
what figure represents the proper diagram of the post-reaction mixture.
Execute the plan:
6 molecules N 2 ×
2 molecules NH 3
= 12 molecules NH 3
1 molecules N 2
2 molecules NH 3
12 molecules H 2 ×
= 8 molecules NH 3
3 molecules H 2
Only 8 molecules NH3 are formed, because the limiting reactant is H2. The excess reactant is N2.
Determine the number of molecules of N2 that actually reacted:
Chapter 4: Quantities of Reactants and Products
12 molecules H 2 ×
167
1 molecules N2
= 4 molecules N 2 reacted
3 molecules H 2
The molecules of N2 left = 6 molecules of N2 initial – 4 molecules of N2 reacted = 2 molecules of N2 left
These results indicate that Figure 4 is the right figure, and that (b) is the right statement regarding the
limiting reactant: “H2 is the limiting reactant”
Check your answers: There are two molecules of N2 left after all the molecules of H2 react, so it makes
sense that H 2 is the limiting reactant. The figure shows 8 molecules of NH3 and 2 molecules of N2. These
answers make sense.
91. The reactants are six O2 molecules and six CO molecules. The given equation described s 1:2 reaction, so we
determine that the CO molecules run out first, after producing six CO2 molecules and using three O2
molecules. That leaves three O2 molecule unreacted:
Note: The black circles are carbon, and the white circles are oxygen.
92. Four XY3 molecules are made from two diatomic X2 molecules and six diatomic Y2 molecules. So,
symbolically, the reaction is 2 X2 + 6 Y2
reaction is (b) X2 + 3 Y2
4 XY3, and the stoichiometric equation representing that
2 XY3
93. The increase of NaHCO3 causes an obvious increase in the amount of gas produced, as indicated by the
larger volume of the balloon in #1 through #4. In #5 and #6, the volume look unchanged, but there are
subtle differences in the shape and “droopiness” of the balloon, indicating a possible increase in the
pressure inside the balloon. If we use volume alone, then the NaHCO3 limits up to #4, then the acetic acid is
the limiting reactant in #5 and #6. If we consider an increase in pressure as an indication of more gas being
produced in each balloon, then we would conclude that the NaHCO3 was limiting in every experiment.
94. When masses smaller than 1.2 gram of the metal are added, the metal is the limiting reactant, so the mass of
the compound produced is directly proportional to the mass of metal present (shown by the straight line
rising up to the right on the graph). More metal makes more products when the mass of metal is less than
1.2 gram.
When the mass larger than 1.2 g of metal are added, the bromine is the limiting reactant, so the particular
mass of the metal is independent of how much compound is made. Since the amount of bromine is held
constant, the mass of compound formed is also a constant (shown by a horizontal line on the graph).
Conclusion: When metal mass is less than 1.2 g the metal is the limiting reactant. When the metal mass is
greater than 1.2 g the bromine is the limiting reactant.
168
Chapter 4: Quantities of Reactants and Products
95. Define the problem: Given the graph of a product’s mass vs. a metal element reactant’s mass, the identity of
the elemental reactants and the information that one compound is formed, determine the empirical formula of
the product compound, write the balanced equation for the reaction, and name the product.
Develop a plan: The point where the graph levels out is the stoichiometric equivalence point. (See the
answer to Question 92 in the Student Solution Manual for details). Assuming 100 % yield, subtract the
metal mass from the product mass to get the other reactant’s mass. Use the molar mass of the reactant
elements to find their moles. Then use a whole-number mole ratio to determine the empirical formula. Write
and balance the equation and name the product.
Execute the plan: Estimating the location of the stoichiometric equivalence point, we find that 12.6 grams of
compound are made from 2.3 grams of Fe.
12.6 g compound – 2.3 g Fe = 10.3 g Br
The compound is a contains only Fe and Br: FeiBrj.
2.3 g Fe ×
1 mol Fe
= 0.041 mol Fe
55.845 mol Fe
10 .3 g Br ×
1 mol Br
= 0.13 mol Br
79.904 mol Br
Set up mole ratio and simplify by dividing by the smallest number of moles:
0.041 mol Fe : 0.13 mol Br
1 Fe : 3 Br
Use the whole number ratio for the subscripts in the formula. The empirical formula is FeBr3. Remembering
that bromine is a diatomic liquid element and that iron is a monatomic solid, the equation looks like this: 2
Fe(s) + 3 Br2(l)
2 FeBr3(s)
The compound is iron(III) bromide.
Check your answers: The approximations that had to be made from the graph provide a wide rage of values,
depending on a persons estimations.. However, the result is reasonable, due to the fact that iron has two
typical oxidation states, +2 and +3, and that bromine has only one, –1. There is no way the data could
support the answer being iron(II) bromide.
More Challenging Problems
96. Define the problem: Given the formulas of the reactants and one product and the percent mass of elements
in the second product, determine the balanced chemical equation.
Develop a plan: Choose a convenient sample mass of product, Fex Oy , such as 100.0 g. Find the mass of Fe
and O in the sample, using the given mass percent. Use the molar mass of the elements to find their moles,
then use a whole-number mole ratio to determine the empirical formula. Using the formulas of the reactants
and products balance the equation.
Execute the plan: 100.0 g of .Fex Oy contains 72.3 g Fe and 27.7 g O.
72 .3 g Fe ×
1 mol Fe
= 1.29 mol Fe
55.845 g Fe
27 .7 g O ×
1 mol O
= 1.73 mol O
15.9994 g O
Set up mole ratio and simplify by dividing by the smallest number of moles:
1.29 mol Fe : 1.73 mol O
Chapter 4: Quantities of Reactants and Products
169
1 Fe : 1.34 O
Multiply by 3 to get whole numbers:
3 Fe : 4 O
Use the whole number ratio for the subscripts in the formula. The empirical formula is Fe3 O4 .
? H2(g) + ? Fe2O3(s)
? H2O(l) + ? Fe3 O4 (s)
Select order: Fe, O, H
? H2(g) + 3 Fe2O3(s)
? H2O(l) + 2 Fe3 O4 (s)
6 Fe
? H2(g) + 3 Fe2O3(s)
1 H2O(l) + 2 Fe3 O4 (s)
9O
1 H2(g) + 3 Fe2O3(s)
1 H2O(l) + 2 Fe3 O4 (s)
2H
Check your answer: Fe3 O4 is a common oxide of iron. 6 Fe, 9 O and 2 H on each side.
97. Define the problem: Given the percent mass of elements in an organic compound, determine the balanced
chemical equation for the combustion reaction.
Develop a plan: Choose a convenient sample mass of product, Cx Hy Oz, such as 100.00 g. Find the mass of
C and H in the sample, using the given mass percent, then subtract those masses from the total sample mass
to get the mass of O. Use the molar mass of the elements to find their moles, then use a whole-number mole
ratio to determine the empirical formula. For combustion, use the formula of the organic compound and O2
as reactants and CO2 and H2 O as products, then balance the equation.
Execute the plan:
100.00 g of .Cx Hy Oz contains 34.62 g C and 3.88 g H.
100.00 g of .Cx Hy Oz – 34.62 g C – 3.88 g H = 61.50 g O
34 .62 g C×
1 mol C
= 2.882 mol C
12.0107 g C
61 .50 g O ×
3.88 g H ×
1 mol H
= 3.85 mol H
1.0079 g H
1 mol O
= 3.844 mol O
15.9994 g O
Set up mole ratio and simplify by dividing by the smallest number of moles:
2.882 mol C : 3.85 mol H : 3.844 mol O
1 C : 1.336 H : 1.334 O
Multiply by 3 to get whole numbers:
3C:4H:4O
Use the whole number ratio for the subscripts in the formula. The empirical formula is C3 H4 O4 .
? C3 H4 O4 + ? O2
? CO2 + ? H2O
Select order: C, H, O
1 C3 H4 O4 + ? O2
3 CO2 + ? H2O
3C
1 C3 H4 O4 + ? O2
3 CO2 + 2 H2O
4H
170
Chapter 4: Quantities of Reactants and Products
1 C3 H4 O4 + 2 O2
3 CO2 + 2 H2O
8O
Check your answer: The mole ratio is quite close to whole number values. 3 C, 4 H, and 8 O on each side.
98. Define the problem: Given a balanced chemical equation, the volume of a liquid limiting reactant, and the
density of the liquid, determine the maximum theoretical yield of a product.
Develop a plan: Use the density of the reactant to find the grams of the reactant. Then use the molar mass
of the reactant to find the moles of reactant. Then use the stoichiometry of the equation to determine moles
of product. Then use the molar mass of the product to get the mass of the product.
Execute the plan:
3.1023 g Br 2
1 mol Br2
1 mol Al2 Br6 533.387 g Al2 Br6
25.0 mL Br2 ×
×
×
×
= 86.3 g Al2Br6
1 mL Br2
159.808 g Br2
3 mol Br2
1 mol Al2 Br6
Check your answer: The units all cancel, and the numerical multipliers are larger than the dividers. This
somewhat larger answer makes sense.
99. Define the problem: Given the formula of two binary oxide compounds both containing an unknown
element and an unknown amount of oxygen, and given the mass of the both elements in a given sample of
each compound, we need to determine the ratio of oxygen atoms in the two formulas. Given the number of
oxygen atoms in one formula, determine the identity of the unknown element.
Develop a plan: We have incomplete information in this problem, so we’ll let the ratio help us eliminate the
need to know information common between the two substances. Start by converting the mass of O in each
sample to moles of O. We then use the mass of A and its molar mass to determine moles of A, then use the
formula stoichiometry to determine the moles of AOx. Dividing the moles of O by the moles of AOx tells us
the value of x. Follow the same procedure to get y. We don’t know molar mass of A, but it will cancel in the
ratio.
Execute the plan:
(a) Find moles of O:
3.2 g O ×
1 mol O
= 0.20 mol O
15.9994 g O
Find moles of AOx, using M A for the molar mass of unknown element A:
1.2 g A ×
1 mol A 1 mol AOx 1.2
×
=
mol AOx
MA g A
1 mol A
MA
Find moles of AOy in a similar fashion:
2.4 g A ×
1 mol A 1 mol AOy
2.4
×
=
mol AO y
MA g A
1 mol A
MA
mol O
x
Now set up the ratio to get :
y
mol AOx
mol O
mol AOy
Chapter 4: Quantities of Reactants and Products
171
0.20 mol O
1.2
mol AOx
MA
2 mol O
1 mol AOx
=
0.20 mol O
1 mol O
2.4
1
mol AO y
mol AOy
MA
x 2
=
y 1
(b) If x = 2, we can use the moles of O to tell use how many moles of A are present in the AOx sample.
1 mol A
0.20 mol O×
= 0.10 mol A
2 mol O
Divide mass of A in the sample by moles of A in the sample to get molar mass of A:
1.2 g A
= 12 g / mol
0.10 mol A
The element with this molar mass is carbon, C.
Check your answers: (a) The law of multiple proportions assures us that the ratio of elements in binary
compounds is a small whole number, so the ratio looks right. There are probably many other legitimate
ways to solve this problem. (b) The molar mass of the unknown element was very close to the molar mass
of carbon, and both CO and CO2 are known compounds, so this result makes sense also.
100.
Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the
limiting reactant, the mass of the product produced, and the mass remaining of the excess reactant when
the reaction is complete.
Develop a plan: (a) Use the molar mass of the reactants to find the moles of the reactant substances.
Then use the stoichiometry of the equation to determine the moles of the product produced in each case.
Identify the limiting reactant from the reactant that produces the least number of products. (b) Use the
moles of product produced from the limiting reactant and the molar mass of the product to find the grams.
(c) From the quantity of limiting reactant, determine the moles of the other reactant needed for complete
reaction. Convert that number to grams using the molar mass. Then subtract the quantity used from the
initial mass given to get the mass of excess reactant.
Execute the plan: Note that 500 grams has 1 significant figure and 1300 grams has 2 significant figures.
(a) The balanced equation says: 1 mol CH4 produces 3 mol H2.
500 g CH 4 ×
1 mol CH4
3 mol H 2
×
= 90 mol H 2
16.0423 g CH 4 1 mol CH 4
The balanced equation says: 1 mol H2O produces 3 mol H2.
1300 g H 2O ×
1 mol H 2O
3 mol H 2
×
= 220 mol H 2
18.0152 g H 2O 1 mol H2 O
The number of H2 moles produced from CH4 is smaller (90 mol < 220 mol), so CH4 is the limiting reactant
and H2 O is the excess reactant.
(b) Find the mass H2:
90 mol H 2 ×
2.0158 g H2
= 200 g H 2
1 mol H 2
Chapter 4: Quantities of Reactants and Products
172
500 g CH 4 ×
(c)
1 mol CH4
1 mol H2O 18.0152 g H 2O
×
×
= 600 g H 2O
16.0423 g CH 4 1 mol CH 4
1 mol H 2O
1300 g H2 O initial – 600 g H2 O used up = 700 g H2 O remains unreacted
Check your answers: The small number of significant figures makes the uncertainty in this calculation
somewhat high. However, the general expectations are still met. There is a larger mass of H2O present, and
the molar masses of the reactants are about the same size. Since the equation indicates that equal moles of
H2O and CH4 react, it makes sense that CH4 is the limiting reactant. In addition, the calculation in (c) proves
that the initial mass of H2 O was larger than required to react with all of the CH4 .
101. Define the problem: Given the mass of a sample of ore with unknown amounts of two copper compounds
and some amo unt of inert impurity, the products of a chemical reaction, and the mass and percent purity
of one element in the products, determine the percent by mass of one of the copper compounds.
Develop a plan: Use the percent purity of the copper in the product and the molar mass of copper to find
the moles of copper in the ore. Set two variables X = the mass in grams of Cu 2 S and Y = the mass in
grams of CuS. Then establish two equations, one describing the total mass of the sample related to the
three components and one describing the moles of copper in the sample from two sources. Algebraically
solve for X and Y, then use the mass of Cu 2 S and the total sample mass to find the percentage by mass.
Execute the plan: 150 .8 g impure Cu ×
90 .0 g Cu
1 mol Cu
×
= 2.136 mol Cu
100 g impure Cu 63.546 g Cu
Let X = grams of Cu 2 S and Y = grams of CuS, and recall that 10.% of the ore is inert impurities.
Total mass of sample = 200.0 g = X + Y + 0.10 × 200.0 g
Moles of copper =
1 mol Cu 2S
2 mol Cu
1 mol CuS
1 mol Cu
2.136 mol Cu = X g Cu 2S ×
×
+ Y g CuS×
×
159.157 g Cu 2S 1 mol Cu 2S
95.611 g CuS 1 mol CuS
180.0 = X + Y
Solve for X:
and
2.136 = 0.0125662 X + 0.010459 Y
2.136 = 0.0125662 X + 0.010459 (180.0 − X)
2.136 = 0.0125662 X + 1.883 − 0.010459 X
2.136 − 1.883 = (0.0125662 − 0.010459) X
0.253 = (0.002107) X
120. = X
120 . g Cu 2 O
× 100 % = 60.1 % Cu 2O
200 .0 g ore
Check your answer: Close to the maximum quantity of iron was produced, so it makes sense that the
percent yield is over ninety percent.
102.
Define the problem: Given the names of products and reactants of a chemical reaction, and the masses of
the two reactants, determine the masses of all the products and reactants after the reaction is complete.
Chapter 4: Quantities of Reactants and Products
173
Develop a plan: Determine the formulas of the reactants and products and set up and balance the
chemical equation. Use the molar mass of the reactants to find the moles of the reactant substances.
Then use the stoichiometry of the equation to determine the moles of the product produced in each case.
Identify the limiting reactant from the reactant that produces the least number of products. Use the moles
of product produced from the limiting reactant, the molar masses of the products to find the grams of
products. Use the moles of product produced, determine the moles of the other reactant needed for
complete reaction. Convert that number to grams using the molar mass. Then subtract the quantity used
from the initial mass given to get the mass of excess reactant.
Execute the plan: Silver nitrate is AgNO3 . Sodium carbonate is Na 2 CO3 . Silver carbonate is Ag2 CO3 .
Sodium nitrate is NaNO3 .
? AgNO3 + ? Na 2 CO3
? Ag 2 CO3 + ? NaNO3
Order: Ag, Na, N, C, O
2 AgNO3 + ? Na 2 CO3
1 Ag2 CO3 + ? NaNO3
2 Ag
2 AgNO3 + 1 Na 2 CO3
1 Ag2 CO3 + 2 NaNO3
2 Na, 2 N, 1 C, 9 O
The balanced equation says: 1 mol Na 2 CO3 produces 1 mol Ag2 CO3 .
1 mol Na 2CO3
1 mol Ag2CO3
12.43 g Na 2 CO3 ×
×
= 0.1173 mol Ag2CO 3
105.9885 g Na 2CO 3 1 mol Na 2 CO 3
The balanced equation says: 2 mol AgNO3 produces 1 mol Ag2 CO3 .
8.37 g AgNO 3 ×
1 mol AgNO 3
1 mol Ag2CO 3
×
= 0.0246 mol Ag2CO 3
169.8731 g AgNO 3 2 mol AgNO 3
The numb er of Ag2 CO3 moles produced from AgNO3 is smaller (0.0246 mol < 0.1173 mol), so AgNO3 is
the limiting reactant and Na 2 CO3 is the excess reactant. Therefore, at the end of the reaction, the mass of
AgNO3 present is zero grams.
Find the mass Ag2 CO3 : 0.0246 mol Ag2CO 3 ×
Find the mass NaNO3 : 0.0246 mol Ag2CO 3 ×
275.7453 g Ag2CO 3
= 6.79 g Ag2 CO3
1 mol Ag 2CO 3
2 mol NaNO 3 84.9947 g NaNO 3
×
= 4.19 g NaNO 3
1 mol Ag 2CO 3
1 mol NaNO 3
Find the mass Na 2 CO3 used up, then subtract from intial for mass unreacted:
1 mol Na 2CO 3 105.9885 g Na 2CO 3
0.0246 mol Ag2CO 3 ×
×
= 2.61 g Na 2CO 3 used up
1 mol Ag 2CO 3
1 mol Na 2CO 3
12.43 g Na 2 CO3 initial – 2.61 g Na 2 CO3 used up = 9.82 g Na 2 CO3 remains unreacted
Check your answers: The total mass before the reaction = 12.43 g + 8.37 g = 20.80 g is equal to the total
mass after the reaction = 6.79 g + 4.19 g + 9.82 g = 20.80 g, in accordance with the conservation of mass.
103. Define the problem: Given a balanced equation for a chemical reaction, and the masses of the three
reactants, determine the limiting reactant, the mass of one of the product and the masses of the excess
reactants after the reaction is complete.
174
Chapter 4: Quantities of Reactants and Products
Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances. Then
use the stoichiometry of the equation to determine the moles of the product produced in each case.
Identify the limiting reactant from the reactant that produces the least number of products. Use the moles
of product produced and the molar mass of the product to find the grams of products. Use the moles of
product produced, determine the moles of the other reactants needed for complete reaction. Convert
those numbers to grams using the molar masses. Then subtract the quantity used from the initial mass
given to get the mass of excess reactant.
Execute the plan:
The balanced equation says: 2 mol SO2 produces 2 mol H2 SO4 .
1 mol SO 2
2 mol H2 SO 4
200. g SO 2 ×
×
= 3.12 mol H 2 SO 4
64.064 g SO 2
2 mol SO 2
The balanced equation says: 1 mol O2 produces 2 mol H2 SO4 .
85 g O 2 ×
1 mol O2
2 mol H2SO 4
×
= 5.3 mol H 2 SO 4
31.9988 g O2
1 mol O 2
The balanced equation says: 1 mol H2 O produces 2 mol H2 SO4 .
1 mol H2 O
2 mol H2SO 4
66 g H2 O×
×
= 3.7 mol H 2 SO 4
18.0152 g H 2 O
2 mol H 2O
The number of H2 SO4 moles produced from SO2 is smaller (3.12 mol < 3.7 mol < 5.3 mol), so SO2 is the
limiting reactant.
98.078 g H 2SO 4
Find the mass H2 SO4 : 3.12 mol H2 SO 4 ×
= 306 g H2SO 4
1 mol H2SO 4
Find the mass H2 O used up, then subtract from intial for mass unreacted:
1 mol SO 2
2 mol H 2 O 18.0152 g H 2O
200. g SO 2 ×
×
×
= 56.2 mol H 2O used up
64.064 g SO 2 2 mol SO 2
1 mol H 2O
66 g H2 O initial – 56.2 g H2 O used up = 10. g H2 O remains unreacted
Find the mass H2 O used up, then subtract from intial for mass unreacted:
200. g SO 2 ×
1 mol SO 2
1 mol O2 31.9988 g O2
×
×
= 49.9 mol O2 used up
64.064 g SO 2 2 mol SO 2
1 mol O2
85 g O2 initial – 49.9 g O2 used up = 35 g O2 remains unreacted
Check your answers: The total mass before the reaction = 200. g + 85 g + 66 g = 351 g is equal to the total
mass after the reaction = 306 g + 10. g + 35 g = 351 g, in accordance with the conservation of mass.
104.
Define the problem: Given the mass of a sample of a liquid containing unknown amounts of two organic
compounds and given the mass of one of the products of a chemical reaction, determine the composition
of the liquid.
Chapter 4: Quantities of Reactants and Products
175
Develop a plan: First, balance the combustion reactions for ethyl alcohol and methyl alcohol. Set two
variables X = the mass in grams of methyl alcohol and Y = the mass in grams of ethyl alcohol. Then
establish two equations, one describing the total mass of the sample related to the two components and
one describing the moles of carbon dioxide produced when burning the sample. Algebraically solve for X
and Y, then use those masses and the total sample mass to find the percentage by mass of the two
compounds.
Execute the plan:
Unbalanced:
Unbalanced:
? C2H5OH(l) + ? O2(g)
? CO2(g) + ? H2O(l) Order: C, H, then O
1 C2H5OH(l) + ? O2(g)
2 CO2(g) + ? H2O(l)
2 C’s
1 C2H5OH(l) + ? O2(g)
2 CO2(g) + 3 H2O(l)
6 H’s
1 C2H5OH(l) + 6 O2(g)
2 CO2(g) + 3 H2O(l)
7 O’s
? CH3OH(l) + ? O2(g)
? CO2(g) + ? H2O(l) Order: C, H, then O
1 CH3OH(l) + ? O2(g)
1 CO2(g) + ? H2O(l)
1 C’s
1 CH3OH(l) + ? O2(g)
1 CO2(g) + 2 H2O(l)
4 H’s
1 CH3OH(l) +
3
O2(g)
2
1 CO2(g) + 2 H2O(l)
4 O’s
Let X = grams of ethyl alcohol and Y = grams of methyl alcohol
Total mass of sample = 0.280 g = X + Y
Moles of carbon dioxide produced when burned = 0.385 g CO2 ×
= X g C2 H5OH ×
1 mol CO2
= 0.00875 mol CO2
44.0095 g CO2
1 mol C 2H 5OH
2 mol CO2
×
46.0682 g C 2H 5OH 1 mol C 2H 5OH
+ Y g CH 3OH ×
0.280 = X + Y
Solve for X:
and
1 mol CH 3OH
1 mol CO 2
×
32.0417 g CH 3OH 1 mol CH 3OH
0.00875 = 0.0434139 X + 0.0312093 Y
0.00875 = 0.0434139 X + 0.0312093(0.280 – X)
0.00875 = 0.0434139 X + 0.000874 − 0.0312093 X
0.00875 − 0.000874 = (0.0434139 − 0.0312093) X
0.00001 = 0.0122046 X
X = 0.0008 g ethyl alcohol
Y = 0.280 – X = 0.280 – 0.0008 = 0.279 g methyl alcohol
0.279 g CH 3OH
×100 % = 99 .7 % CH 3OH
0.280 g liquid
0.001 g C2 H5OH
× 100 % = 0.3 % C2 H 5OH
0.280 g liquid
176
Chapter 4: Quantities of Reactants and Products
Check your answer: Assuming the liquid is pure methyl alcohol gives the following mass of CO2 :
0.280 g CH 3OH ×
1 mol CH 3OH
1 mol CO 2
44.0095 g CO 2
×
×
= 0.385 g CO 2
32.0417 g CH 3OH 1 mol CH 3OH
1 mol CO 2
So it makes sense that the liquid is almost one hundred percent methanol.
105. Define the problem: Given the percent mass of elements in an organic compound, determine the balanced
chemical equation for the combustion reaction.
Develop a plan: Choose a convenient sample mass of product, Cx Hy Oz, such as 100.00 g. Find the mass
of C and H in the sample, using the given mass percent, then subtract those masses from the total sample
mass to get the mass of O. Use the molar mass of the elements to find their moles, then use a wholenumber mole ratio to determine the empirical formula. For combustion, use the formula of the organic
compound and O2 as reactants and CO2 and H2 O as products, then balance the equation.
Execute the plan:
100.00 g of .Cx Hy Oz contains 49.31 g C and 6.90 g H.
100.00 g of .Cx Hy Oz – 49.31 g C – 6.90 g H = 43.79 g O
49.31 g C×
1 mol C
= 4.106 mol C
12.0107 g C
43.79 g O ×
6.90 g H ×
1 mol H
= 6.85 mol H
1.0079 g H
1 mol O
= 2.737 mol O
15 .9994 g O
Set up mole ratio and simplify by dividing by the smallest number of moles:
4.106 mol C : 6.85 mol H : 2.737 mol O
1.5 C : 2.5 H : 1 O
Multiply by 2 to get whole numbers:
3C:5H:2O
Use the whole number ratio for the subscripts in the formula. The empirical formula is C3 H5 O2 .
? C3 H5 O2 + ? O2
? CO2 + ? H2O
Select order: H, C, O
2 C3 H5 O2 + ? O2
? CO2 + 5 H2O
10 H
2 C3 H5 O2 + ? O2
6 CO2 + 5 H2O
6C
2 C3 H5 O2 +
13
O2
2
4 C3 H5 O2 + 13 O2
6 CO2 + 5 H2O
17 O
12 CO2 + 10 H2O
Check your answer: The mole ratio is quite close to whole number values. 12 C, 20 H, and 34 O on each
side.
106. Define the problem: Given the percent mass of elements in an organic compound and the compounds
molar mass, determine the empirical formula and the molecular formula.
Chapter 4: Quantities of Reactants and Products
177
Develop a plan: Choose a convenient sample mass of product, Cx Hy NzOw, such as 100.00 g. Find the
mass of C and H in the sample, using the given mass percent, then subtract those masses from the total
sample mass to get the mass of O. Use the molar mass of the elements to find their moles, then use a
whole-number mole ratio to determine the empirical formula. For combustion, use the formula of the
organic compound and O2 as reactants and CO2 and H2 O as products, then balance the equation.
Execute the plan:
(a) 100.00 g of .Cx Hy NzOw contains 54.82 g C, 7.10 g N, and 32.46 g O.
100.00 g of .Cx Hy NzOw – 54.82 g C – 7.10 g N – 32.46 g O = 5.62 g H
54 .82 g C×
1 mol C
= 4.564 mol C
12.0107 g C
32 .46 g O ×
1 mol O
= 2.029 mol O
15.9994 g O
7.10 g N ×
1 mol N
= 0.507 mol N
14.0067 g N
5.62 g H ×
1 mol H
= 5.58 mol H
1.0079 g H
Set up mole ratio and simplify by dividing by the smallest number of moles:
4.564 mol C : 5.58 mol H : 0.507 mol N : 2.029 mol O
9 C : 11 H : 1 N : 4 O
Use the whole number ratio for the subscripts in the formula. The empirical formula is C9 H11 NO4 .
(b) The molar mass of the empirical formula = 197.1875 g/mol C9 H11 NO4
197 .19 g / mol compound
= 1 emp . formula / compound
197 .1875 g / mol emp . formula
The molecular formula is C9 H11 NO4 .
Check your answers: The mole ratio is quite close to whole number values. The empirical formula is very
close to the molecular formula. These answers make sense.