Download Ch 9 Book Answer

Chapter 9
Chemical Quantities
1.
Although we define mass as the “amount of matter in a substance,” the units in which we
measure mass are a human invention. Atoms and molecules react on an individual particleby-particle basis, and we have to count individual particles when doing chemical calculations.
2.
Balanced chemical equations tell us in what proportions on a mole basis substances combine;
since the molar masses of C(s) and O2(g) are different, 1 g of O2 could not represent the same
number of moles as 1 g of C.
3.
a.
2NO(g) + O2(g) 2NO2(g)
Two molecules of nitrogen monoxide combine with one molecule of oxygen gas,
producing two molecules of nitrogen dioxide. Two moles of gaseous nitrogen
monoxide combine with one mole of gaseous oxygen, producing two moles of
gaseous nitrogen dioxide.
b.
2AgC2H3O2(aq) + CuSO4(aq) Ag2SO4(s) + Cu(C2H3O2)2(aq)
Note: The term “formula unit” is used in the following statement because the
substances involved in the above reaction are ionic, and do not contain true
molecules. Two formula units of silver acetate will react with one formula unit of
copper(II) sulfate, precipitating one formula unit of silver sulfate and leaving one
formula unit of copper(II) acetate in solution. Two moles of aqueous silver acetate
react with one mole of aqueous copper(II) sulfate, to produce one mole of solid silver
sulfate as a precipitate, and leaving one mole of copper(II) acetate in solution.
c.
PCl3(l) + 3H2O(l) H3PO3(l) + 3HCl(g)
One molecule of phosphorus trichloride reacts with three molecules of water,
producing one molecule of phosphorous acid and three molecules of gaseous
hydrogen chloride. One mole of liquid phosphorus trichloride reacts with three moles
of liquid water, producing one mole of liquid phosphorous acid and three moles of
gaseous hydrogen chloride.
d.
C2H6(g) + Cl2(g) C2H5Cl(g) + HCl(g)
One molecule of ethane (C2H6) reacts with one molecule of chlorine, producing one
molecule of chloroethane (C2H5Cl) and one molecule of hydrogen chloride. One
mole of gaseous ethane combines with one mole of chlorine gas, giving one mole of
gaseous chloroethane and one mole of gaseous hydrogen chloride.
4.
a.
3MnO2(s) + 4Al(s) 3Mn(s) + 2Al2O3(s)
Three formula units of manganese(IV) oxide react with four aluminum atoms,
producing three manganese atoms and two formula units of aluminum oxide. Three
moles of solid manganese(IV) oxide react with four moles of solid aluminum, to
produce three moles of solid manganese and two moles of solid aluminum oxide.
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Chemical Quantities
b.
75
B2O3(s) + 3CaF2(s) 2BF3(g) + 3CaO(s)
One molecule of diboron trioxide reacts with three formula units of calcium fluoride,
producing two molecules of boron trifluoride and three formula units of calcium
oxide. One mole of solid diboron trioxide reacts with three moles of solid calcium
fluoride, to give two moles of gaseous boron trifluoride and three moles of solid
calcium oxide.
c.
3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)
Three molecules of nitrogen dioxide [nitrogen(IV) oxide] react with one molecule of
water, to produce two molecules of nitric acid and one molecule of nitrogen
monoxide [nitrogen(II) oxide]. Three moles of gaseous nitrogen dioxide react with
one mole of liquid water, to produce two moles of aqueous nitric acid and one mole
of nitrogen monoxide gas.
d.
C6H6(g) + 3H2(g) C6H12(g)
One molecule of C6H6 (which is named benzene) reacts with three molecules of
hydrogen, producing just one molecule of C6H12 (which is named cyclohexane). One
mole of gaseous benzene reacts with three moles of hydrogen gas, giving one mole of
gaseous cyclohexane.
5.
False. The coefficients of the balanced chemical equation represent the ratios on a mole basis
by which hydrogen peroxide decomposes.
6.
For converting from a given number of moles of CH4 to the number of moles of oxygen
needed for reaction, the correct mole ratio is
2 mol O 2
1 mol CH 4
For converting from a given number of moles of CH4 to the number of moles of product
produced, the ratios are
1 mol CO 2
1 mol CH 4
7.
2 mol H 2 O
1 mol CH 4
2Ag(s) + H2S(g) Ag2S(s) + H2(g)
1 mol Ag 2S
2 mol Ag
8.
and
and
1 mol H 2
2 mol Ag
a. 2FeO(s) + C(s) 2Fe(l) + CO2(g)
0.125 mol FeO 0.125 mol FeO World of Chemistry
2 mol Fe
2 mol FeO
1 mol CO 2
2 mol FeO
= 0.125 mol Fe
= 0.0625 mol CO2
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76
Chapter 9
b.
Cl2(g) + 2KI(aq) 2KCl(aq) + I2(s)
0.125 mol KI 0.125 mol KI c.
2 mol KCl
1 mol I 2
2 mol KI
0.125 mol Na2B4O7 0.125 mol CaC2 a.
4 mol H 3 BO3
1 mol Na 2 B4 O7
1 mol Na 2SO 4
1 mol Na 2 B4 O7
= 0.500 mol H3BO3
= 0.125 mol Na2SO4
CaC2(s) + 2H2O(l) Ca(OH)2(s) + C2H2(g)
0.125 mol CaC2 9.
= 0.0625 mol I2
Na2B4O7(s) + H2SO4(aq) + 5H2O(l) 4H3BO3(s) + Na2SO4(aq)
0.125 mol Na2B4O7 d.
= 0.125 mol KCl
2 mol KI
1 mol Ca(OH) 2
1 mol CaC 2
1 mol C 2 H 2
1 mol CaC 2
= 0.125 mol Ca(OH)2
= 0.125 mol C2H2
NH3(g) + HCl(g) NH4Cl(s)
molar mass of NH4Cl, 53.49 g
0.50 mol NH3 1 mol NH 4 Cl
1 mol NH 3
0.50 mol NH4Cl b.
= 0.50 mol NH4Cl
53.49 g NH 4 Cl
1 mol NH 4 Cl
= 27 g NH4Cl
CH4(g) + 4S(g) CS2(l) + 2H2S(g)
molar masses: CS2, 76.15 g; H2S, 34.09 g
0.50 mol S 1 mol CS2
4 mol S
0.125 mol CS2 0.50 mol S c.
76.15 g CS2
1 mol CS2
2 mol H 2S
4 mol S
0.25 mol H2S = 0.125 mol CS2 (= 0.13 mol CS2)
= 9.5 g CS2
= 0.25 mol H2S
34.09 g H 2S
1 mol H 2S
= 8.5 g H2S
PCl3(l) + 3H2O(l) H3PO3(aq) + 3HCl(aq)
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Chemical Quantities
77
molar masses: H3PO3, 81.99 g; HCl, 36.46 g
0.50 mol PCl3 1 mol H 3 PO3
81.99 g H 3 PO3
0.50 mol H3PO3 0.50 mol PCl3 1.50 mol HCl d.
= 0.50 mol H3PO3
1 mol PCl3
= 41 g H3PO3
1 mol H 3 PO3
3 mol HCl
1 mol PCl3
= 1.5 mol HCl
36.46 g HCl
1 mol HCl
= 54.7 = 55 g HCl
NaOH(s) + CO2(g) NaHCO3(s)
molar mass of NaHCO3, 84.01 g
1 mol NaHCO3
0.50 mol NaOH 1 mol NaOH
0.50 mol NaHCO3 10.
= 0.50 mol NaHCO3
84.01 g NaHCO3
1 mol NaHCO3
= 42 g NaHCO3
Before doing the calculations, the equations must be balanced.
a.
4KO2(s) + 2H2O(l) 3O2(g) + 4KOH(s)
3 mol O 2
0.625 mol KOH b.
3 mol Se
2 mol H 2 O
= 0.938 mol Se
2CH3CH2OH(l) + O2(g) 2CH3CHO(aq) + 2H2O(l)
0.625 mol H2O d.
= 0.469 mol O2
SeO2(g) + 2H2Se(g) 3Se(s) + 2H2O(g)
0.625 mol H2O c.
4 mol KOH
2 mol CH 3CHO
2 mol H 2 O
= 0.625 mol CH3CHO
Fe2O3(s) + 2Al(s) 2Fe(l) + Al2O3(s)
0.625 mol Al2O3 2 mol Fe
1 mol Al2 O3
= 1.25 mol Fe
11.
the molar mass of the substance
12.
Stoichiometry is the process of using a chemical equation to calculate the relative masses of
reactants and products involved in a reaction.
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13.
Chapter 9
a.
molar mass of Ag = 107.9 g
1 mol
2.01 10–2 g Ag b.
molar mass of (NH4)2S = 68.15 g
45.2 mg (NH4)2S c.
10 g
a.
1 kg
= 81.6 mol SO2
1 mol
241.9 g
= 1.12 mol Fe(NO3)3
1g
1000 mg
1 mol
151.9 g
= 8.36 10–5 mol FeSO4
molar mass of LiOH = 23.95 g
1 mol
23.95 g
= 288.5 = 289 mol LiOH
molar mass of CaCO3 = 100.1 g
100.1 g
1 mol
= 0.0221 g CaCO3
molar mass of He = 4.003 g
4.003 g
1 mol
= 11.0 g He
molar mass of O2 = 32.00 g
0.00975 mol O2 d.
64.07 g
molar mass of FeSO4 = 151.9 g
2.75 mol He c.
1 mol
molar mass of Fe(NO3)3 = 241.9 g
2.21 10–4 mol CaCO3 b.
= 6.63 10–4 mol (NH4)2S
= 2.59 10–7 mol U
238.0 g
1000 g
6.91 103 g LiOH 14.
68.15 g
molar mass of SO2 = 64.07 g
12.7 mg FeSO4 g.
1 mol
1 mol
6
272 g Fe(NO3)3 f.
1000 mg
1g
5.23 kg SO2 e.
1g
molar mass of uranium = 238.0 g
61.7 g U d.
= 1.86 10–4 mol Ag
107.9 g
32.00 g
1 mol
= 0.312 g O2
molar mass of CO2 = 44.01 g
7.21 millimol = 0.00721 mol
0.00721 mol CO2 World of Chemistry
44.01 g
1 mol
= 0.317 g CO2
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Chemical Quantities
e.
79
molar mass of FeS = 87.92 g
87.92 g
0.835 mol FeS f.
molar mass of KOH = 56.11 g
56.11 g
4.01 mol KOH g.
= 225 g KOH
1 mol
molar mass of H2 = 2.016 g
2.016 g
0.0219 mol H2 15.
= 73.4 g FeS
1 mol
= 0.0442 g H2
1 mol
Before any calculations are done, the equations must be balanced.
a.
Mg(s) + CuCl2(aq) MgCl2(aq) + Cu(s)
molar mass of Mg = 24.31 g
25.0 g Mg 1 mol
1.03 mol Mg b.
= 1.03 mol Mg
24.31 g
1 mol CuCl2
1 mol Mg
= 1.03 mol CuCl2
2AgNO3(aq) + NiCl2(aq) 2AgCl(s) + Ni(NO3)2(aq)
molar mass of AgNO3 = 169.9 g
25.0 g AgNO3 1 mol
169.9 g
0.147 mol AgNO3 c.
= 0.147 mol AgNO3
1 mol NiCl2
2 mol AgNO3
= 0.0735 mol NiCl2
NaHSO3(aq) + NaOH(aq) Na2SO3(aq) + H2O(l)
molar mass of NaHSO3 = 104.1 g
25.0 g NaHSO3 1 mol
104.1 g
0.240 mol NaHSO3 d.
= 0.240 mol NaHSO3
1 mol NaOH
1 mol NaHSO3
= 0.240 mol NaOH
KHCO3(aq) + HCl(aq) KCl(aq) + H2O(l) + CO2(g)
molar mass of KHCO3 = 100.1 g
25.0 g KHCO3 1 mol
100.1 g
0.250 mol KHCO3 World of Chemistry
= 0.250 mol KHCO3
1 mol HCl
1 mol KHCO3
= 0.250 mol HCl
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80
16.
Chapter 9
Before any calculations are done, the equations must be balanced. Since the given and
required quantities in this question are shown in milligrams, it is most convenient to perform
the calculations in terms of millimoles of the substances involved. One millimole of a
substance represents the molar mass of the substance expressed in milligrams.
a.
FeSO4(aq) + K2CO3(aq) FeCO3(s) + K2SO4(aq)
millimolar masses: FeSO4, 151.9 mg; FeCO3, 115.9 mg; K2SO4, 174.3 mg
1 mmol FeSO 4
10.0 mg FeSO4 0.0658 mmol FeSO4 0.0658 mmol FeSO4 b.
= 0.0658 mmol FeSO4
151.9 mg FeSO 4
1 mmol FeCO3
1 mmol FeSO 4
1 mmol K 2SO 4
1 mmol FeSO 4
115.9 mg FeCO3
1 mmol FeCO3
174.3 mg K 2SO 4
1 mmol K 2SO 4
= 7.63 mg FeCO3
= 11.5 mg K2SO4
4Cr(s) + 3SnCl4(l) 4CrCl3(s) + 3Sn(s)
millimolar masses: Cr, 52.00 mg; CrCl3, 158.4 mg; Sn, 118.7 mg
10.0 mg Cr 1 mmol Cr
52.00 mg Cr
4 mmol CrCl3
0.192 mmol Cr 4 mmol Cr
3 mmol Sn
0.192 mmol Cr c.
= 0.192 mmol Cr
4 mmol Cr
158.4 mg CrCl3
1 mmol CrCl3
118.7 mg Sn
1 mmol Sn
= 30.4 mg CrCl3
= 17.1 mg Sn
16Fe(s) + 3S8(s) 8Fe2S3(s)
millimolar masses: S8, 256.6 mg; Fe2S3, 207.9 mg
10.0 mg S8 1 mmol S8
256.6 mg S8
8 mmol Fe 2S3
0.0390 mmol S8 d.
= 0.0390 mmol S8
3 mmol S8
207.9 mg Fe 2S3
1 mmol Fe 2S3
= 21.6 mg Fe2S3
3Ag(s) + 4HNO3(aq) 3AgNO3(aq) + 2H2O(l) + NO(g)
millimolar masses:
HNO3, 63.0 mg; AgNO3, 169.9 mg
H2O, 18.0 mg; NO, 30.0 mg
10.0 mg HNO3 1 mmol HNO3
63.0 mg HNO3
0.159 mmol HNO3 World of Chemistry
= 0.159 mmol HNO3
3 mmol AgNO 3
4 mmol HNO 3
169.9 mg AgNO 3
1 mmol AgNO 3
= 20.3 mg AgNO3
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Chemical Quantities
81
0.159 mmol HNO3 0.159 mmol HNO3 17.
2 mmol H 2 O
4 mmol HNO 3
1 mmol NO
4 mmol HNO 3
18.0 mg H 2 O
1 mmol H 2 O
30.0 mg NO
1 mmol NO
= 1.43 mg H2O
= 1.19 mg NO
2H2(g) + O2(g) 2H2O(g)
molar masses: H2, 2.016 g; H2O, 18.02 g
56.0 g H2 1 mol H 2
2.016 g H 2
27.77 mol H2 2 mol H 2 O
= 27.77 mol H2O
2 mol H 2
18.02 g H 2 O
27.77 mol H2O 18.
= 27.77 mol H2
1 mol H 2 O
= 500. g H2O
2H2(g) + O2(g) 2H2O(g)
molar masses of O2 = 32.00 g
1 mol O 2
0.0275 mol H2 2 mol H 2
32.00 g O 2
0.01375 mol O2 19.
= 0.01375 = 0.0138 mol O2
1 mol O 2
= 0.440 g O2
molar masses: C, 12.01 g; CO, 28.01 g; CO2, 44.01 g
5.00 g C 1 mol C
12.01 g C
= 0.4163 mol C
carbon dioxide: C(s) + O2(g) CO2(g)
0.4163 mol C 1 mol CO 2
= 0.4163 mol CO2
1 mol C
0.4163 mol CO2 44.01 g CO 2
1 mol CO 2
= 18.3 g CO2
carbon monoxide: 2C(s) + O2(g) 2CO(g)
0.4163 mol C 2 mol CO
0.4163 mol CO World of Chemistry
2 mol C
= 0.4163 mol CO
28.01 g CO
1 mol CO
= 11.7 g CO
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20.
Chapter 9
2Fe(s) + 3Cl2(g) 2FeCl3(s)
millimolar masses: Fe, 55.85 mg; FeCl3, 162.2 mg
15.5 mg Fe 1 mmol Fe
55.85 mg Fe
2 mmol FeCl3
0.2775 mmol Fe = 0.2775 mmol FeCl3
2 mmol Fe
0.2775 mmol FeCl3 21.
= 0.2775 mmol Fe
162.2 mg FeCl3
1 mmol FeCl3
= 45.0 mg FeCl3
2H2O2(aq) 2H2O(l) + O2(g)
molar masses: H2O2, 34.02 g; O2, 32.00 g
10.00 g H2O2 1 mol H 2 O 2
0.2939 mol H2O2 0.1470 mol O2 22.
= 0.2939 mol H2O2
34.02 g H 2 O 2
1 mol O 2
2 mol H 2 O 2
32.00 g O 2
= 0.1470 mol O2
= 4.704 g O2
1 mol O 2
Cu(s) + S(s) CuS(s)
molar masses: Cu, 63.55 g; S, 32.07 g
1.25 g Cu 1 mol
= 1.97 10–2 mol Cu
63.55 g
1.97 10–2 mol Cu 1.97 10–2 mol S 23.
1 mol S
1 mol Cu
32.07 g
1 mol
= 1.97 10–2 mol S
= 0.631 g S
2NH4NO3(s) 2N2(g) + O2(g) + 4H2O(g)
molar masses: NH4NO3, 80.05 g; N2, 28.02 g; O2, 32.00 g; H2O, 18.02 g
1.25 g NH4NO3 1 mol NH 4 NO3
80.05 g NH 4 NO3
0.0156 mol NH4NO3 0.0156 mol N2 World of Chemistry
= 0.0156 mol NH4NO3
2 mol N 2
2 mol NH 4 NO3
28.02 g N 2
1 mol N 2
= 0.0156 mol N2
= 0.437 g N2
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Chemical Quantities
83
1 mol O 2
0.0156 mol NH4NO3 0.00780 mol O2 2 mol NH 4 NO3
32.00 g O 2
= 0.250 g O2
1 mol O 2
4 mol H 2 O
0.0156 mol NH4NO3 2 mol NH 4 NO3
18.02 g H 2 O
0.0312 mol H2O = 0.00780 mol O2
= 0.0312 mol H2O
= 0.562 g H2O
1 mol H 2 O
As a check, note that 0.437 g + 0.250 g + 0.562 g = 1.249 g = 1.25 g.
24.
2Mg(s) + O2(g) 2MgO(s)
molar masses: Mg, 24.31 g; MgO, 40.31 g
1.25 g Mg 1 mol
24.31 g
= 5.14 10–2 mol Mg
5.14 10–2 mol Mg 2 mol MgO
5.14 10–2 mol MgO 25.
= 5.14 10–2 mol MgO
2 mol Mg
40.31 g
= 2.07 g MgO
1 mol
From the balanced equation:
Cl2 + 2KI I2 + 2KCl
We can calculate the following:
4.50 103 g Cl2 26.
1 mol Cl2
70.9 g Cl2
1 mol I 2
1 mol Cl2
253.8 g I 2
1 mol I 2
= 1.61 104 g I2
From the balanced equation:
Cl2 + F2 2ClF
We can calculate the following:
5.00 10–3 g ClF 27.
1 mol ClF
54.45 g ClF
1 mol Cl2
2 mol ClF
70.9 g Cl2
1 mol Cl2
= 3.26 10–3 g Cl2
Start by balancing the equations:
C3H8 + 5O2 3CO2 + 4H2O
2C4H10 + 13O2 8CO2 + 10H2O
If we had, for example, 10.00 g of C3H8 and 10.00 g of C4H10, we can calculate the mass of
oxygen required for each reaction and compare the results.
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84
Chapter 9
10.00 g C3H8 10.00 g C4H10 1 mol C3 H 8
44.09 g C3 H 8
1 mol C 4 H 10
5 mol O 2
1 mol C3 H 8
13 mol O 2
58.12 g C 4 H 10
2 mol C 4 H 10
32.00 g O 2
1 mol O 2
= 36.29 g O2
32.00 g O 2
1 mol O 2
= 35.79 g O2
Therefore, more O2 would be required for the combustion of C3H8. The same result is
obtained with masses other than 10.00 g.
28.
Start with the balanced equations:
CH4 + 2O2 CO2 + 2H2O
4NH3 + 5O2 4NO + 6H2O
The amount of water produced from 1.00 g CH4 is:
1.00 g CH4 1 mol CH 4
16.042 g CH 4
2 mol H 2 O
1 mol CH 4
= 0.125 mol H2O
The mass of NH3 needed to produce 0.125 mol H2O is:
0.125 mol H2O 4 mol NH 3
6 mol H 2 O
17.034 g NH 3
1 mol NH 3
= 1.42 g NH3
29.
The limiting reactant is the reactant that limits the amounts of products that can form in a
chemical reaction. All given reactants are necessary for the production of products: if the
limiting reactant has been consumed, then there is none of this reactant present for reaction.
30.
We start with 6 molecules N2 and 6 molecules H2, and these react according to the balanced
equation:
N2 + 3H2 2NH3
We can determine amounts of product and leftover reactant from this information given.
N2 + 3H2 2NH3
start
react
end
6
6
–2
–6
4
0
0
+4
4
Note that these react in the same ratio as given in the balanced equation (that is,
2:6:4 = 1:3:2). The pictures should show 4 molecules N2 and 4 molecules NH3.
31.
To determine the limiting reactant, first calculate the number of moles of each reactant
present. Then determine how these numbers of moles correspond to the stoichiometric ratio
indicated by the balanced chemical equation for the reaction.
32.
The theoretical yield of a reaction represents the stoichiometric amount of product that should
form if the limiting reactant for the process is completely consumed.
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Chemical Quantities
85
33.
A reactant is present in excess if there is more of that reactant present than is needed to
combine with the limiting reactant for the process. By definition, the limiting reactant cannot
be present in excess. An excess of any reactant does not affect the theoretical yield for a
process: the theoretical yield is determined by the limiting reactant.
34.
a.
2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g)
Molar masses: Al, 26.98 g; HCl, 36.46 g; AlCl3, 133.3 g; H2, 2.016 g
15.0 g Al 1 mol
26.98 g
15.0 g HCl = 0.556 mol Al
1 mol
36.46 g
= 0.411 mol HCl
Since HCl is needed to react with Al in a 6:2 (i.e., 3:1) molar ratio, it seems pretty
certain that HCl is the limiting reactant. To prove this, we can calculate the quantity
of Al that would react with the given number of moles of HCl.
0.411 mol HCl 2 mol Al
6 mol HCl
= 0.137 mol Al
By this calculation we have shown that all the HCl present will be needed to react
with only 0.137 mol Al (out of the 0.556 mol Al present). Therefore, HCl is the
limiting reactant, and Al is present in excess. The calculation of the masses of
products produced is based on the number of moles of the limiting reactant.
0.411 mol HCl 0.411 mol HCl b.
2 mol AlCl3
6 mol HCl
3 mol H 2
6 mol HCl
133.3 g
1 mol
2.016 g
1 mol
= 18.3 g AlCl3
= 0.414 g H2
2NaOH(aq) + CO2(g) Na2CO3(aq) + H2O(l)
molar masses: NaOH, 40.00 g; CO2, 44.01 g; Na2CO3, 105.99 g; H2O, 18.02 g
15.0 g NaOH 15.0 g CO2 1 mol
40.00 g
1 mol
44.01 g
= 0.375 mol NaOH
= 0.341 mol CO2
For the 0.375 mol NaOH, let's calculate if there is enough CO2 present to
react:
0.375 mol NaOH World of Chemistry
1 mol CO 2
2 mol NaOH
= 0.1875 mol CO2 (0.188 mol)
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86
Chapter 9
We have present more CO2 (0.341 mol) than is needed to react with the given
quantity of NaOH. NaOH is therefore the limiting reactant, and CO2 is present in
excess. The quantities of products resulting are based on the complete conversion of
the limiting reactant (0.375 mol NaOH):
1 mol Na 2 CO3
0.375 mol NaOH 2 mol NaOH
1 mol H 2 O
0.375 mol NaOH c.
2 mol NaOH
105.99 g
1 mol
18.02 g
1 mol
= 19.9 g Na2CO3
= 3.38 g H2O
Pb(NO3)2(aq) + 2HCl(aq) PbCl2(s) + 2HNO3(aq)
Molar masses: Pb(NO3)2, 331.2 g; HCl, 36.46 g; PbCl2, 278.1 g;
HNO3, 63.02 g
15.0 g Pb(NO3)2 15.0 g HCl 1 mol
331.2 g
1 mol
36.46 g
= 0.0453 mol Pb(NO3)2
= 0.411 mol HCl
With such a large disparity between the numbers of moles of the reactants, it’s
probably a sure bet that Pb(NO3)2 is the limiting reactant. To confirm this, we can
calculate how many mol of HCl are needed to react with the given amount of
Pb(NO3)2.
0.0453 mol Pb(NO3)2 2 mol HCl
1 mol Pb(NO3 ) 2
= 0.0906 mol HCl
We have considerably more HCl present than is needed to react completely with the
Pb(NO3)2. Therefore, Pb(NO3)2 is the limiting reactant, and HCl is present in excess.
The quantities of products produced are based on the limiting reactant being
completely consumed.
0.0453 mol Pb(NO3)2 0.0453 mol Pb(NO3)2 d.
1 mol PbCl2
1 mol Pb(NO 3 ) 2
2 mol HNO 3
1 mol Pb(NO 3 ) 2
278.1 g
1 mol
63.02 g
1 mol
= 12.6 g PbCl2
= 5.71 g HNO3
2K(s) + I2(s) 2KI(s)
Molar masses: K, 39.10 g; I2, 253.8 g; KI, 166.0 g
15.0 g K 15.0 g I2 World of Chemistry
1 mol
39.10 g
1 mol
253.8 g
= 0.384 mol K
= 0.0591 mol I2
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Chemical Quantities
87
Since, from the balanced chemical equation, we need twice as many moles of K as
moles of I2, and since there is so little I2 present, it’s a safe bet that I2 is the limiting
reactant. To confirm this, we can calculate how many moles of K are needed to react
with the given amount of I2 present:
2 mol K
0.0591 mol I2 1 mol I 2
= 0.1182 mol K
Clearly we have more potassium present than is needed to react with the small
amount of I2 present. I2 is therefore the limiting reactant, and potassium is present in
excess. The amount of product produced is calculated from the number of moles of
the limiting reactant present:
2 mol KI
0.0591 mol I2 35.
a.
1 mol I 2
166.0 g
= 19.6 g KI
1 mol
2NH3(g) + 2Na(s) 2NaNH2(s) + H2(g)
Molar masses: NH3, 17.03 g; Na, 22.99 g; NaNH2, 39.02 g
50.0 g NH3 50.0 g Na 1 mol
= 2.94 mol NH3
17.03 g
1 mol
22.99 g
= 2.17 mol Na
Since the coefficients of NH3 and Na are the same in the balanced chemical equation
for the reaction, the two reactants combine in a 1:1 molar ratio. Therefore, Na is the
limiting reactant, which will control the amount of product produced.
2.17 mol Na b.
2 mol NaNH 2
2 mol Na
39.02 g
1 mol
= 84.7 g NaNH2
BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq)
Molar masses: BaCl2, 208.2 g; Na2SO4, 142.1 g; BaSO4, 233.4 g
50.0 g BaCl2 1 mol
208.2 g
50.0 g Na2SO4 = 0.240 mol BaCl2
1 mol
142.1 g
= 0.352 mol Na2SO4
Since the coefficients of BaCl2 and Na2SO4 are the same in the balanced chemical
equation for the reaction, the reactant having the smaller number of moles present
(BaCl2) must be the limiting reactant, which will control the amount of product
produced.
0.240 mol BaCl2 World of Chemistry
1 mol BaSO 4
1 mol BaCl2
233.4 g
1 mol
= 56.0 g BaSO4
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88
Chapter 9
c.
SO2(g) + 2NaOH(aq) Na2SO3(aq) + H2O(l)
Molar masses: SO2, 64.07 g; NaOH, 40.00 g; Na2SO3, 126.1 g
50.0 g SO2 1 mol
64.07 g
50.0 g NaOH = 0.780 mol SO2
1 mol
40.00 g
= 1.25 mol NaOH
From the balanced chemical equation for the reaction, every time one mol of SO2
reacts, two mol of NaOH is needed. For 0.780 mol of SO2, 2(0.780 mol) = 1.56 mol
of NaOH would be needed. We do not have sufficient NaOH to react with the SO2
present: therefore, NaOH is the limiting reactant, and controls the amount of product
obtained.
1.25 mol NaOH d.
1 mol Na 2SO 3
2 mol NaOH
126.1 g
1 mol
= 78.8 g Na2SO3
2Al(s) + 3H2SO4(l) Al2(SO4)3(s) + 3H2(g)
Molar masses: Al, 26.98 g; H2SO4, 98.09 g; Al2(SO4)3, 342.2 g
50.0 g Al 1 mol
26.98 g
50.0 g H2SO4 = 1.85 mol Al
1 mol
98.09 g
= 0.510 mol H2SO4
Since the amount of H2SO4 present is smaller than the amount of Al, let’s see if H2SO4 is
the limiting reactant by calculating how much Al would react with the given amount of
H2SO4.
0.510 mol H2SO4 2 mol Al
3 mol H 2SO 4
= 0.340 mol Al
Since all the H2SO4 present would react with only a small portion of the Al present,
H2SO4 is therefore the limiting reactant and will control the amount of product obtained.
0.510 mol H2SO4 36.
1 mol Al2 (SO 4 )3
3 mol H 2SO 4
342.2 g
1 mol
= 58.2 g Al2(SO4)3
a. CO(g) + 2H2(g) CH3OH(l)
CO is the limiting reactant; 11.4 mg CH3OH
b. 2Al(s) + 3I2(s) 2AlI3(s)
I2 is the limiting reactant; 10.7 mg AlI3
c. Ca(OH)2(aq) + 2HBr(aq) CaBr2(aq) + 2H2O(l)
HBr is the limiting reactant; 12.4 mg CaBr2; 2.23 mg H2O
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Chemical Quantities
89
d. 2Cr(s) + 2H3PO4(aq) 2CrPO4(s) + 3H2(g)
H3PO4 is the limiting reactant; 15.0 mg CrPO4; 0.309 mg H2
37.
Cl2(g) + 2NaI(aq) 2NaCl(aq) + I2(s)
Br2(l) + 2NaI(aq) 2NaBr(aq) + I2(s)
molar masses: Cl2, 70.90 g; Br2, 159.8 g; I2, 253.8 g; NaI, 149.9 g
1 mol Cl2
5.00 g Cl2 70.90 g Cl2
25.0 g NaI = 0.0705 mol Cl2
1 mol NaI
= 0.167 mol NaI
149.9 g NaI
Since we would need 2(0.0705) = 0.141 mol of NaI for the Cl2 to react completely, and we
have more than this amount of NaI, then Cl2 must be the limiting reactant.
0.0705 mol Cl2 0.0705 mol I2 1 mol I 2
1 mol Cl2
253.8 g I 2
1 mol I 2
1 mol Br2
5.00 g Br2 159.8 g Br2
25.0 g NaI = 0.0705 mol I2
= 17.9 g I2
= 0.0313 mol Br2
1 mol NaI
149.9 g NaI
= 0.167 mol NaI
Since we would need 2(0.0313) = 0.0626 mol of NaI for the Br2 to react completely, and we
have more than this amount of NaI, then Br2 must be the limiting reactant.
0.0313 mol Br2 0.0313 mol I2 38.
1 mol I 2
1 mol Br2
253.8 g I 2
1 mol I 2
= 0.0313 mol I2
= 7.94 g I2
4Fe(s) + 3O2(g) 2Fe2O3(s)
Molar masses: Fe, 55.85 g; Fe2O3, 159.7 g
1.25 g Fe 1 mol
55.85 g
= 0.0224 mol Fe present
Calculate how many mol of O2 are required to react with this amount of Fe.
0.0224 mol Fe World of Chemistry
3 mol O 2
4 mol Fe
= 0.0168 mol O2
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90
Chapter 9
Since we have more O2 than this, Fe must be the limiting reactant.
2 mol Fe 2 O3
0.0224 mol Fe 4 mol Fe
0.0112 mol Fe2O3 39.
= 0.0112 mol Fe2O3
159.7 g Fe 2 O3
1 mol Fe 2 O3
= 1.79 g Fe2O3
Ca2+(aq) + Na2C2O4(aq) CaC2O4(s) + 2Na+(aq)
molar masses: Ca2+, 40.08 g; Na2C2O4, 134.0 g
2+
15 g Ca 1 mol Ca 2+
= 0.37 mol Ca2+
40.08 g Ca 2+
1 mol Na 2 C 2 O 4
15 g Na2C2O4 134.0 g Na 2 C 2 O 4
= 0.11 mol Na2C2O4
Since the balanced chemical equation tells us that one oxalate ion is needed to precipitate
each calcium ion, from the number of moles calculated to be present, it should be clear that
not nearly enough sodium oxalate ion has been added to precipitate all the calcium ion in the
sample.
40.
The actual yield for a reaction is the quantity of product actually isolated from the reaction
vessel. The theoretical yield represents the mass of products that should be produced by the
reaction if the limiting reactant is fully consumed. The percent yield represents what fraction
of the amount of product that should have been collected was actually collected.
41.
If the reaction is performed in a solvent, the product may have a substantial solubility in the
solvent; the reaction may come to equilibrium before the full yield of product is achieved;
loss of product may occur through operator error.
42.
Percent yield =
43.
S8(s) + 8Na2SO3(aq) + 40H2O(l) 8Na2S2O3•5H2O
actual yield
theoretical yield
100 =
1.23 g
1.44 g
100 = 85.4%
molar masses: S8, 256.6 g; Na2SO3, 126.1 g; Na2S2O3•5H2O, 248.2 g
3.25 g S8 1 mol S8
256.6 g S8
13.1 g Na2SO3 = 0.01267 mol S8
1 mol Na 2SO3
126.1 g Na 2SO3
= 0.1039 mol Na2SO3
S8 is the limiting reactant.
0.01267 mol S8 World of Chemistry
8 mol Na 2S2 O3 • 5H 2 O
1 mol S8
= 0.1014 mol Na2S2O3•5H2O
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Chemical Quantities
91
248.2 g Na 2S2 O3 • 5H 2 O
0.1014 mol Na2S2O3•5H2O actual yield
Percent yield =
44.
= 25.2 g Na2S2O3•5H2O
1 mol Na 2S2 O3 • 5H 2 O
100 =
theoretical yield
5.26 g
100 = 20.9%
25.2 g
2LiOH(s) + CO2(g) Li2CO3(s) + H2O(g)
molar masses: LiOH, 23.95 g; CO2, 44.01 g
1 mol LiOH
155 g LiOH 23.95 g LiOH
1 mol CO 2
2 mol LiOH
44.01 g CO 2
1 mol CO 2
= 142 g CO2
Since the cartridge has only absorbed 102 g CO2 out of a total capacity of 142 g CO2, the
cartridge has absorbed
102 g
142 g
45.
100 = 71.8% of its capacity
Xe(g) + 2F2(g) XeF4(s)
molar masses: Xe, 131.3 g; F2, 38.00 g; XeF4, 207.3 g
130. g Xe 100. g F2 1 mol Xe
= 0.9901 mol Xe
131.3 g Xe
1 mol F2
38.00 g F2
= 2.632 mol F2
Xe is the limiting reactant.
0.9901 mol Xe 1 mol XeF4
1 mol Xe
0.9901 mol XeF4 Percent yield =
46.
= 0.9901 mol XeF4
207.3 g XeF4
1 mol XeF4
actual yield
theoretical yield
= 205 g XeF4
100 =
145 g
205 g
100 = 70.7%
CaCO3(s) + 2HCl(g) CaCl2(s) + CO2(g) + H2O(g)
molar masses: CaCO3, 100.1 g; HCl, 36.46 g; CaCl2, 111.0 g
155 g CaCO3 250. g HCl 1 mol CaCO3
100.1 g CaCO3
1 mol HCl
36.46 g HCl
= 1.548 mol CaCO3
= 6.857 mol HCl
CaCO3 is the limiting reactant.
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92
Chapter 9
1.548 mol CaCO3 1.548 mol CaCl2 Percent yield =
47.
1 mol CaCl2
= 1.548 mol CaCl2
1 mol CaCO3
111.0 g CaCl2
= 172 g CaCl2
1 mol CaCl2
actual yield
100 =
theoretical yield
142 g
172 g
100 = 82.6%
NaCl(aq) + NH3(aq) + H2O(l) + CO2(s) NH4Cl(aq) + NaHCO3(s)
molar masses: NH3, 17.03 g; CO2, 44.01 g; NaHCO3, 84.01 g
10.0 g NH3 15.0 g CO2 1 mol NH 3
17.03 g NH 3
1 mol CO 2
44.01 g CO 2
= 0.5872 mol NH3
= 0.3408 mol CO2
CO2 is the limiting reactant.
1 mol NaHCO3
0.3408 mol CO2 1 mol CO 2
0.3408 mol NaHCO3 48.
= 0.3408 mol NaHCO3
84.01 g NaHCO3
1 mol NaHCO3
= 28.6 g NaHCO3
Fe(s) + S(s) FeS(s)
molar masses: Fe, 55.85 g; S, 32.07 g; FeS, 87.92 g
5.25 g Fe 12.7 g S 1 mol Fe
55.85 g Fe
1 mol S
32.07 g S
= 0.0940 mol Fe
= 0.396 mol S
Fe is the limiting reactant.
0.0940 mol Fe 49.
1 mol FeS
1 mol Fe
87.92 g FeS
1 mol FeS
= 8.26 g FeS produced
C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g)
molar masses: glucose, 180.2 g; CO2, 44.01 g
1.00 g glucose 1 mol glucose
180.2 g glucose
5.549 10–3 mol glucose 3.33 10–2 mol CO2 World of Chemistry
= 5.549 10–3 mol glucose
6 mol CO 2
1 mol glucose
44.01 g CO 2
1 mol CO 2
= 3.33 10–2 mol CO2
= 1.47 g CO2
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Chemical Quantities
50.
93
10.3 g Cl
–
mass of Cl present = 1.054 g sample 100.0 g sample
= 0.1086 g Cl–
molar masses: Cl–, 35.45 g; AgNO3, 169.9 g; AgCl, 143.4 g
0.1086 g Cl– 1 mol Cl
= 3.063 10–3 mol Cl–
35.45 g Cl
3.063 10–3 mol Cl– 1 mol AgNO 3
1 mol Cl
3.063 10–3 mol AgNO3 3.063 10–3 mol Cl– 51.
For O2:
52.
a.
5 mol O 2
1 mol C3 H 8
= 0.520 g AgNO3 required
= 3.063 10–3 mol AgCl
143.4 g AgCl
1 mol AgCl
For CO2:
= 0.439 g AgCl produced
3 mol CO 2
1 mol C3 H 8
For H2O:
4 mol H 2 O
1 mol C3 H 8
2H2O2(l) 2H2O(l) + O2(g)
0.50 mol H2O2 2 mol H 2 O
2 mol H 2 O 2
1 mol O 2
2 mol H 2 O 2
= 0.50 mol H2O
= 0.25 mol O2
2KClO3(s) 2KCl(s) + 3O2(g)
0.50 mol KClO3 0.50 mol KClO3 c.
1 mol AgNO3
1 mol Cl-
0.50 mol H2O2 b.
169.9 g AgNO3
1 mol AgCl
3.063 10–3 mol AgCl = 3.063 10–3 mol AgNO3
2 mol KCl
2 mol KClO3
3 mol O 2
2 mol KClO3
= 0.50 mol KCl
= 0.75 mol O2
2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g)
0.50 mol Al 0.50 mol Al World of Chemistry
2 mol AlCl3
2 mol Al
3 mol H 2
2 mol Al
= 0.50 mol AlCl3
= 0.75 mol H2
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94
Chapter 9
d.
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
3 mol CO 2
0.50 mol C3H8 1 mol C3 H 8
4 mol H 2 O
0.50 mol C3H8 53.
a.
= 1.5 mol CO2
1 mol C3 H 8
= 2.0 mol H2O
NH3(g) + HCl(g) NH4Cl(s)
molar mass of NH3 = 17.0 g
1 mol NH 3
1.00 g NH3 17.0 g NH 3
0.0588 mol NH3 b.
= 0.0588 mol NH3
1 mol NH 4 Cl
1 mol NH 3
= 0.0588 mol NH4Cl
CaO(s) + CO2(g) CaCO3(s)
molar mass CaO = 56.1 g
1.00 g CaO 1 mol CaO
56.1 g CaO
0.0178 mol CaO c.
= 0.0178 mol CaO
1 mol CaCO3
1 mol CaO
= 0.0178 mol CaCO3
4Na(s) + O2(g) 2Na2O(s)
molar mass Na = 22.99 g
1.00 g Na 1 mol Na
0.0435 mol Na d.
= 0.0435 mol Na
22.99 g Na
2 mol Na 2 O
4 mol Na
= 0.0217 mol Na2O
2P(s) + 3Cl2(g) 2PCl3(l)
molar mass P = 30.97 g
1.00 g P 1 mol P
30.97 g P
0.0323 mol P 54.
= 0.0323 mol P
2 mol PCl3
2 mol P
= 0.0323 mol PCl3
2Na2O2(s) + 2H2O(l) 4NaOH(aq) + O2(g)
molar masses: Na2O2, 77.98 g; O2, 32.00 g
3.25 g Na2O2 World of Chemistry
1 mol Na 2 O 2
77.98 g Na 2 O 2
= 0.0417 mol Na2O2
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Chemical Quantities
95
0.0417 mol Na2O2 0.0209 mol O2 55.
1 mol O 2
2 mol Na 2 O 2
32.00 g O 2
= 0.0209 mol O2
= 0.669 g O2
1 mol O 2
2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)
150 g = 1.5 102 g
molar masses: C2H2, 26.04 g; O2, 32.00 g
1 mol C 2 H 2
1.5 102 g C2H2 5.760 mol C2H2 14.40 mol O2 56.
a.
26.04 g C 2 H 2
5 mol O 2
2 mol C 2 H 2
32.00 g O 2
1 mol O 2
= 5.760 mol C2H2
= 14.40 mol O2
= 4.6 102 g O2
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
molar masses: C2H5OH, 46.07 g; O2, 32.00 g; CO2, 44.01 g
1 mol C 2 H 5OH
25.0 g C2H5OH 25.0 g O2 46.07 g C 2 H 5OH
1 mol O 2
32.00 g O 2
= 0.5427 mol C2H5OH
= 0.7813 mol O2
Since there is less C2H5OH present on a mole basis, see if this substance is the
limiting reactant.
0.5427 mol C2H5OH 3 mol O 2
1 mol C 2 H 5OH
= 1.6281 mol O2
From the above calculation, C2H5OH must not be the limiting reactant (even though
there is a smaller number of moles of C2H5OH present) since more oxygen than is
present would be required to react completely with the C2H5OH present. Oxygen is
the limiting reactant.
0.7813 mol O2 2 mol CO 2
0.5209 mol CO2 b.
3 mol O 2
= 0.5209 mol CO2
44.01 g CO 2
1 mol CO 2
= 22.9 g CO2
N2(g) + O2(g) 2NO(g)
molar masses: N2, 28.02 g; O2, 32.00 g; NO, 30.01 g
25.0 g N2 World of Chemistry
1 mol N 2
28.02 g N 2
= 0.8922 mol N2
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96
Chapter 9
25.0 g O2 1 mol O 2
= 0.7813 mol O2
32.00 g O 2
Since the coefficients of N2 and O2 are the same in the balanced chemical equation
for the reaction, an equal number of moles of each substance would be necessary for
complete reaction. Since there is less O2 present on a mole basis, O2 must be the
limiting reactant.
0.7813 mol O2 1.5626 mol NO c.
2 mol NO
= 1.5626 mol NO
1 mol O 2
30.01 g NO
1 mol NO
= 46.9 g NO
2NaClO2(aq) + Cl2(g) 2ClO2(g) + 2NaCl(aq)
molar masses: NaClO2, 90.44 g; Cl2, 70.90 g; NaCl, 58.44 g
25.0 g NaClO2 25.0 g Cl2 1 mol NaClO 2
= 0.2764 mol NaClO2
90.44 g NaClO 2
1 mol Cl2
70.90 g Cl2
= 0.3526 mol Cl2
See if NaClO2 is the limiting reactant.
0.2764 mol NaClO2 1 mol Cl2
1 mol NaClO 2
= 0.1382 mol Cl2
Since 0.2764 mol of NaClO2 would require only 0.1382 mol Cl2 to react completely
(and since we have more than this amount of Cl2), then NaClO2 must indeed be the
limiting reactant.
0.2764 mol NaClO2 0.2764 mol NaCl d.
2 mol NaCl
2 mol NaClO 2
58.44 g NaCl
1 mol NaCl
= 0.2764 mol NaCl
= 16.2 g NaCl
3H2(g) + N2(g) 2NH3(g)
molar masses: H2, 2.016 g; N2, 28.02 g; NH3, 17.03 g
25.0 g H2 25.0 g N2 1 mol H 2
2.016 g H 2
1 mol N 2
28.02 g N 2
= 12.40 mol H2
= 0.8922 mol N2
See if N2 is the limiting reactant.
0.8922 mol N2 World of Chemistry
3 mol H 2
1 mol N 2
= 2.677 mol H2
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Chemical Quantities
97
N2 is clearly the limiting reactant, since there is 12.40 mol H2 present (a large
excess).
0.8922 mol N2 1.784 mol NH3 57.
2 mol NH 3
= 1.784 mol NH3
1 mol N 2
17.03 g NH 3
1 mol NH 3
= 30.4 g NH3
N2H4(l) + O2(g) N2(g) + 2H2O(g)
molar masses: N2H4, 32.05 g; O2, 32.00 g; N2, 28.02 g; H2O, 18.02 g
20.0 g N2H4 20.0 g O2 1 mol N 2 H 2
= 0.624 mol N2H4
32.05 g N 2 H 2
1 mol O 2
32.00 g O 2
= 0.625 mol O2
The two reactants are present in nearly the required ratio for complete reaction (due to the 1:1
stoichiometry of the reaction and the very similar molar masses of the substances). We will
consider N2H4 as the limiting reactant in the following calculations.
1 mol N 2
0.624 mol N2H4 0.624 mol N2 1 mol N 2 H 4
28.02 g N 2
1 mol N 2
1.248 mol H2O = 17.5 g N2
2 mol H 2 O
0.624 mol N2H4 = 0.624 mol N2
1 mol N 2 H 4
18.02 g H 2 O
1 mol H 2 O
= 1.248 mol H2O = 1.25 mol H2O
= 22.5 g H2O
40 g actual
58.
12.5 g theoretical 59.
We are concerned with the balanced equation:
100 g theoretical
= 5.0 g
X4 + 12HCl 4XCl3 + 6H2
We are given the mass of X4 (248 g). If we can determine the number of moles of X4, we can
determine its molar mass, and therefore its identity. We can do this from the mass of H2.
24.0 g H2 1 mol H 2
2.016 g H 2
1 mol X 4
6 mol H 2
= 1.98 mol X4
Thus,
248 g X 4
1.98 mol X 4
World of Chemistry
= 125 g/mol (molar mass of X4)
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98
Chapter 9
Element X has an atomic mass of about
125
= 31.3 g/mol.
4
This is closest to phosphorus (30.97 g/mol).
60.
We know the following:
x + y xy
5 g 15 g
x + 3z xz3
3 g 18 g
and
Thus, the mole ratio between x and y is 1:1 and the mole ratio between x and z is 1:3,
respectively.
So, the relative masses of x:y are 1:3, and the relative masses of x:z are 3:6 or 1:2 (if x = 3 g,
3z = 18 g or z = 6 g).
Thus, x:y:z = 1:3:2.
If y = 60 g/mol, x = 20 g/mol, and z = 40 g/mol.
61.
This is a “double” limiting reactant problem. First, determine the maximum yield of P4O10
from the equation:
P4 + 5O2 P4O10
20.0 g P4 30.0 g O2 1 mol P4
123.88 g P4
1 mol O 2
32.00 g O 2
1 mol P4 O10
1 mol P4
1 mol P4 O10
5 mol O 2
283.88 g P4 O10
1 mol P4 O10
283.88 g P4 O10
1 mol P4 O10
= 45.8 g P4O10
= 53.2 g P4O10
Thus, the maximum yield of P4O10 is 45.8 g.
Determine the yield of H3PO4 from the equation:
P4O10 + 6H2O 4H3PO4
45.8 g P4O10 15.0 g H2O 1 mol P4 O10
283.88 g P4 O10
1 mol H 2 O
18.016 g H 2 O
4 mol H 3 PO 4
1 mol P4 O10
4 mol H 3 PO 4
6 mol H 2 O
97.994 g H 3 PO 4
1 mol H 3 PO 4
97.994 g H 3 PO 4
1 mol H 3 PO 4
= 63.2 g H3PO4
= 54.4 g H3PO4
Thus, the maximum yield of H3PO4 is 54.4 g.
62.
a. Neither is limiting.
1 mol N2 2 mol NH 3
1 mol N 2
= 2 moles NH3
3 mol H2 2 mol NH 3
= 2 moles NH3
3 mol H 2
b. H2 is limiting.
3 mol H2 2 mol NH 3
= 2 moles NH3
3 mol H 2
c. H2 is limiting.
3 mol H2 2 mol NH 3
= 2 moles NH3
3 mol H 2
d. So, choice d (each would produce the same amount of product) is the correct answer.
World of Chemistry
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Chemical Quantities
99
63.
Choice “c” is correct. Since the molar mass of O2 is larger than the molar mass of C2H6, an
equal mass of each would result in a fewer number of moles of O2 than C2H6. If this were the
case, O2 must be limiting since it is needed in a greater amount (a 7:2 mol ratio).
64.
50.0 g CaO 1 mol CaO 56.08 g CaO
5 mol C = 2.23 mol C required
2 mol CaO
50.0 g C 1 mol C = 4.16 mol C
12.01 g C
Leftover C = 4.16 mol – 2.23 mol = 1.93 mol C 12.01 g C = 23.2 g C left over
1 mol C
(CaO is limiting.)
65.
13.97 g KClO3 Percent yield =
1 mol KClO3
74.55 g KCl
= 8.498 g KCl
2 mol KCl 122.55 g KClO3
2 mol KClO3
1 mol KCl
actual yield
6.23 g
100% =
100% = 73.3% yield
theoretical yield
8.498 g
66.
False. Amounts used can vary. However, according to the equation, for every 28.02 g of N2
(2 mol) we need 6.048 g of H2 (3 mol). So, for example, if we have 28.02 g of N2 and 5.5 g of
H2, the H2 is limiting. If we have 28.02 g of N2 and 6.5 g of H2, the N2 is limiting.
67.
a.
b.
As we add more sodium, we get more product because sodium is the limiting
reactant. However, eventually sodium is in excess (and chlorine is limiting), so the
amount of product does not increase with an increase in sodium.
58.44 g NaCl
20.0 g Na 1 mol Na 2 mol NaCl = 50.8 g NaCl
22.99 g Na
2 mol Na
1 mol NaCl
c.
If we look at the graph, we can see that 40.0 g of sodium reacts in a stoichiometric
ratio with chlorine
(that is, there is no limiting reactant). Because of this, we can use 40.0 g of sodium to
determine the amount of Cl2. Since the problem states that the mass of Cl2 is the same
in all containers, each container will have this amount of Cl2.
1 mol Cl2
70.9 g Cl2
40.0 g Na 1 mol Na = 61.7 g Cl2
22.99 g Na
d.
1 mol Cl2
We should add “since the amount of product is equal when 40.0 g or 50.0 g of
sodium is used, 50.0 g of sodium must be an excess amount.” Since we know from
part c how much Cl2 is in the container, we use the amount of Cl2 (the limiting
reactant) to determine the amount of product.
61.7 g Cl2 e.
2 mol Na
1 mol Cl2
58.44 g NaCl
= 101.7 g NaCl = 102 g NaCl
2 mol NaCl 70.9 g Cl2
1 mol Cl2
1 mol NaCl
In part b, sodium is limiting, and 30.85 g Cl2 is left over (half of the Cl2 is left over).
In part d, Cl2 is limiting and 10.0 g Na is left over (50.0 g – 40.0 g).
World of Chemistry
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