Download Chemistry Chapter 12

Chemistry Chapter 12
Stoichiometry
12.1 The Arithmetic of Equations
Using Balanced Chemical Equations
-chemists use balanced chemical equations as a basis to calculate how much reactant is needed or
product formed in a reaction
-equations are the recipes that tell chemists what amounts of reactants to mix and what masses of
products to expect
-the calculations of quantities using a chemical reaction equations called stoichiometry
-stoichiometry is a form of chemical bookkeeping
-if you want to increase or decrease the quantity you adjust the amounts of reactants appropriately
Interpreting Chemical Equations
CAN BE INTERPRETED IN TERMS OF ATOMS, MOLECULES, MOLES, MASS, OR
VOLUME
-one interpretation is the number of atoms
indicates that the number and type of each atom that makes up each reactant also makes up
each product
the number and type of atom are not changed during a chemical reaction
-a second interpretation of the coefficients is to indicate particles or molecules
N2(g) + 3 H2(g)  2 NH3(g)
1 molecule of nitrogen gas reactions with three molecules of hydrogen gas to produce two
molecules of ammonia gas
the ratio of molecules of nitrogen gas to molecules of hydrogen gas is 1:3
-a third interpretation of the coefficients is to indicate moles
N2(g) + 3 H2(g)  2 NH3(g)
a mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of
ammonia gas
notice that the total moles of reactants does not equal total moles of products
-a fourth interpretation is the use of mass and the law of conservation of mass
N2(g) + 3 H2(g)  2 NH3(g)
the coefficients cannot stand for mass in grams, as 1 g of N2 does not react with 3 g
H2 gas to produce 2 g of NH3 gas
however, 28 g of nitrogen (1 mole) will react with 6 g of hydrogen gas (3 moles) to
produce 34 g (2 moles) of ammonia gas - mass is conserved
-a fifth interpretation is the use of the coefficients to stand for volumes (in liters)
all reactants and products must be gases
N2(g) + 3 H2(g)  2 NH3(g)
1 L of nitrogen gas will react with 3 L of hydrogen gas to produce 2 L of ammonia gas
this is true since (from Avogadro's hypothesis) equal volumes of all gases under the same
conditions contain the same number of molecules)
notice that volume is not conserved (1 L + 3 L  2 L)
remember the concept of molar volume - that 1 mole of any gas at STP has a volume of
22.4 liters
only mass and atoms are conserved in every chemical reaction
Questions:
1. Interpret the equation below in terms of (a) moles; (b) molecules; (c) volumes; (d) masses
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(g)
(a) _________________________________________________________________________________
(b) __________________________________________________________________________________
(c) __________________________________________________________________________________
(d) __________________________________________________________________________________
2.
2 Fe(s) + 3 Cl2(g)  2 FeCl3(s)
(a) _________________________________________________________________________________
(b) __________________________________________________________________________________
(c) __________________________________________________________________________________
(d) __________________________________________________________________________________
12.2 Chemical Calculations
Writing and Using Mole Ratios
-the most important use of coefficients in a balanced chemical reaction is as moles
-mass which may be given in a problem must be converted to moles
-the ratio of the coefficients is a very important number in equation calculations
N2(g) + 3 H2(g)  2 NH3(g)
-in the above equation the ratio of N2 to H2 is 1:3
-in the above equation the ratio of H2 to NH3 is 3:2
-in the above equation the ratio of N2 to NH3 is 1:2
-an example using this equation is shown below starting with 4 moles of hydrogen as
4 mol H2 2 mol NH3
= 2.67 moles NH3
3 mol H2
-be sure to include units so that dimensional analysis can be used
Sample Problems:
3. Use the equation to the right to do the following calculations using the appropriate mole ratio value be sure to label this value to make sure it is correct.
N2(g) + 3 H2(g)  2 NH3(g)
(a) 2.5 moles of N2 gas will react with how many moles of H2?
_________________________________________________
(b) 3.7 moles of ammonia gas will be produced from how many moles of hydrogen gas?
_________________________________________________
(c) 3.5 moles of hydrogen gas will produce how many moles of ammonia gas?
_________________________________________________
Mass-Mass Calculations
-generally measure mass in grams
-make a conversion from mass of the given to moles of the given
-make a conversion from moles of given to moles of unknown
-make a conversion from moles of unknown to mass of unknown
-see sample problem 12-3 on page 361
-may be done in one complete calculation or in individual steps
-will need to determine the molar mass of the reactants and products when doing problems
The coefficients do not affect the molar mass in stoichiometry calculations.
Example Problem:
The decomposition of potassium chlorate by heat is shown below. What mass of potassium chloride will
be produced from the decomposition of 13.50 g of potassium chlorate.
 2 KCl(s) + 3 O2(g)
2 KClO3(s) heat
_________
______
_________
Step 1: Carefully read the problem. Place any information concerning amount (either grams, mL of
liquid, of liters of a gas) above the reaction equation. Place information concerning solution
concentration (to be studied later in the chapter) below the reaction equation.
Step 2: Determine what quantity is to be found, and place and x above that substance.
Put 13.50 g value above the potassium chlorate and an "x" above the potassium chloride.
Step 3: Use the atomic weight chart to determine the molar mass (to the nearest one-hundredth of a gram)
of each reactant or product which will be used in the problem calculation.
Below the KClO3 put 122.55 g and below the KCl put 74.55 g.
Step 3: Change 13.50 g of potassium chlorate to moles, use the coefficient ratio, then determine
the mass of potassium chloride. This is generally done in one continuous set of calculations.
1350
. g KClO3
1 mole KClO3 2 moles KCl
74.55 g
 8.21 g KCl
122.55 g KClO3 2 moles KClO3 1 mole KCl
4. What mass of oxygen gas will be produced from the decomposition of 8.00 g of potassium chlorate.
 2 KCl(s) + 3 O2(g)
2 KClO3(s) heat
_________
______
_________
________________________________________________________________________________
5. What mass of potassium chlorate will be needed to produce 5.25 g of oxygen gas?
 2 KCl(s) + 3 O2(g)
2 KClO3(s) heat
_________
______
_________
________________________________________________________________________________
Problem Summary Information
-remember that the coefficients in a balanced chemical reaction indicate the ratio of MOLES
of the substances in the compound
-can use this mole relationship to go from one substance to another
-for this calculation it makes no difference if you are working with reactants or products
-when the phrase "an excess of" is used in the wording of the problem, this simply means that
not all of that chemical will be consumed when the reaction is complete
-the answer should have the same number of significant figures as the original quantity given
in the problem - generally 3 significant digits
Sample Problems:
6. What mass of aluminum solid will be needed to react with 16.0 g of nickel(II) sulfate?
2 Al(s) + 3 NiSO4(s)  3 Ni(s) + Al2(SO4)3(s)
________________________________________________________________________________
7. What mass of nickel solid will be produced from the reaction of 5.00 g of aluminum with an excess of
nickel(II) sulfate?
2 Al(s) + 3 NiSO4(s)  3 Ni(s) + Al2(SO4)3(s)
________________________________________________________________________________
8. What mass of nickel(II) sulfate will be needed to produce 25.00 g of nickel solid?
2 Al(s) + 3 NiSO4(s)  3 Ni(s) + Al2(SO4)3(s)
________________________________________________________________________________
Volume-Volume Gas Problems
-if reactants and products are gases, the coefficients can stand for volumes
-if reactants and products under the same conditions, ratios are valid even if the gases are not
at STP
Mass-Volume Problems
-will need to use molar volume value of 22.4 L/mole if working with both solids and gases
-calculations are done only at STP
-do not need to determine molar mass of the gas
Sample Problems:
9. What volume of oxygen gas will be needed to react with 10.0 L of acetylene?
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(g)
____________________________________________________________________________________
10. What volume of carbon dioxide gas will be produced from the complete combustion of 15.0 liters of
acetylene?
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(g)
____________________________________________________________________________________
11. What volume of water vapor will be produced from the complete combustion of 6.25 liters of
acetylene?
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(g)
____________________________________________________________________________________
12. What volume of acetylene gas will be needed to completely consume 8.25 liters of oxygen gas?
2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(g)
____________________________________________________________________________________
13. What volume of oxygen gas will react with 5.00 g of aluminum.
4 Al(s) + 3 O2(g)  2 Al2O3(s)
____________________________________________________________________________________
14. What mass of aluminum will be needed to completely react with 10.0 L of oxygen gas (volume
measured at STP)?
4 Al(s) + 3 O2(g)  2 Al2O3(s)
____________________________________________________________________________________
15. What mass of aluminum oxide will be produced from the reaction of 17.5 liters of oxygen gas with an
excess of aluminum?
4 Al(s) + 3 O2(g)  2 Al2O3(s)
____________________________________________________________________________________
16. What mass of carbon will be needed to produce 15.0 liters of carbon monoxide?
SiO2(s) + 3 C(s)  SiC(s) + 2 CO(g)
___________________________________________________________________________________
17. What mass of silicon dioxide will be needed to produce 10.0 liters of carbon monoxide gas?
SiO2(s) + 3 C(s)  SiC(s) + 2 CO(g)
___________________________________________________________________________________
18. What volume of carbon monoxide will be produced from the reaction of 10.0 g of silicon dioxide
with an excess of carbon?
SiO2(s) + 3 C(s)  SiC(s) + 2 CO(g)
___________________________________________________________________________________
Molarity and Mass-Mass Problems
-many reactions take place in water solutions because of the increase in the mobility of the ions
-most single and double replacement reactions take place in water solutions
-rather than measuring the mass of reactants, the volumes of the solution that reacts is used
-generally use only the molarity unit (M) which has units of moles/liter
Reminders:
-be sure to change all volumes to liters
-do not need molar mass of compounds which are in solution
-put the volume of solution above the reaction equation and the concentration below the equation
Sample Problems
19. What mass of silver will be produced from the complete reaction of 208 mL of a 0.100 M silver
nitrate solution with an excess of copper?
Cu(s) + 2 AgNO3(aq)  Cu(NO3)2(aq) + 2 Ag(s)
____________________________________________________________________________________
20. What volume of 0.215 M silver nitrate solution will be needed to completely react with 3.50 g of
copper?
Cu(s) + 2 AgNO3(aq)  Cu(NO3)2(aq) + 2 Ag(s)
____________________________________________________________________________________
21. What volume of 0.250 M aluminum chloride solution will be needed to completely react with
135 mL of 0.525 M sodium carbonate solution?
2 AlCl3(aq) + 3 Na2CO3(aq)  Al2(CO3)3(s) + 6 NaCl(aq)
_____________________________________________________________________________________
22. What volume of 0.350 M aluminum chloride solution will be required to produce 2.50 g of solid
aluminum carbonate?
2 AlCl3(aq) + 3 Na2CO3(aq)  Al2(CO3)3(s) + 6 NaCl(aq)
_____________________________________________________________________________________
23. What mass of aluminum carbonate will be produced from the reaction of 225 mL of
0.350 M aluminum chloride solution with an excess of sodium carbonate solution.
2 AlCl3(aq) + 3 Na2CO3(aq)  Al2(CO3)3(s) + 6 NaCl(aq)
_____________________________________________________________________________________
12.3 Limiting Reagent and Percent Yield (Reaction Efficiency)
Review:
-balanced chemical reactions can be used to make predictions
-balanced chemical reactions may be written that do not occur
-the equation does not give the exact conditions needed to make the reaction occur
-the equation does not describe the behavior of the atoms during the reaction
What is the Limiting Reactant
-in order for a reaction to occur between atoms, the atoms involved must collide
-with limiting reactants are talking about substances on the left hand side of the reaction equation
-for this type of problem, must be given the quantity of two different reactants
-the reactant that is completely consumed is called the limiting reactant
-this substance is usually the more expensive
-when the reaction is complete, the amount of this substance left is zero
-the limiting reactant is the one which determines what amount of product will be produced
-in the problem set up there are two lines for work since cannot tell which reactant is in excess
and which is the limiting reactant
Sample Problems:
24. If 2.50 g of zinc is dropped into 225 mL of a 2.50 M HCl(aq) solution, what is the reactant in excess
and by what amount?
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
zinc:
_____________________________________________________________________________
HCl:
_____________________________________________________________________________
_______ Reactant in Excess
_______ Amount of Excess
_________ Limiting Reactant
25. If 3.00 g of aluminum is added to 225 mL of 0.250 M copper(II) sulfate solution, what is the reactant
in excess and by what amount?
2 Al(s) + 3 CuSO4(aq)  3 Cu(s) + 2 Al2(SO4)3(aq)
Zn
_______________________________________________________________________________
CuSO4 _______________________________________________________________________________
_______ Reactant in Excess
_______ Amount of Excess
_________ Limiting Reactant
What mass of pure copper will be produced in the above reaction?
_____________________________________________________________________________________
Treat limiting reactant problems as two separate problems. This method, from past experience with
students, seems to be easier to understand than to work with the moles as it done in the text book.
Percent Yield (or Reaction Efficiency)
-the amounts of products actually produced in a chemical reaction may not be the exact amounts
predicted using the reaction equation
-the amounts produced is often less than predicted
-reasons why amount produced is less include:
1. some of the chemicals involved in the reaction may not be completely pure
2. some of the products may be lost during the reaction
3. some of the products may be dissolved in a solution and not able to be recovered
4. some of the reactants may remain - they do not react
-the actual yield is the amount which is produced in a given reaction
-the theoretical yield is the maximum amount of product that could be formed from given
amounts of reactants
-the percent yield cannot exceed 100%
-the percent yield is a measure of the efficiency of a reaction carried out in the laboratory
-see sample problems 12.9 on page 374 and 12.10 on page 375
actual amount of product
-percent yield =
x 100
theoretical amount of product
Problem Solution Hints:
-write quantity of reactant and product above the reaction equation
-put an x above the product - that is what you want to find (do not use value in problem)
-after finding x (the theoretical amount of product) determine the percent yield
-remember that the percent yield is always less than 100%
Sample Problems:
26. When 45.8 g of potassium carbonate reacts with an excess of hydrochloric acid, 36.3 g of potassium
chloride is produced. Determine the percent yield.
K2CO3(s) + 2 HCl(aq)  2 KCl(s) + H2O(l) + CO2(g)
_____________________________________________________________________________________
Percent Yield = ______________ = __________
27. If 6.57 g of iron is reacted with an excess of hydrochloric acid, hydrogen gas and 14.23 g of
iron(II) chloride is produced. Determine the percent yield.
Fe(s) + 2 HCl(aq)  FeCl2(aq) + H2(g)
_____________________________________________________________________________________
Percent Yield = ______________ = __________