CHAPTER
11
SOLUTIONS MANUAL
Stoichiometry
Section 11.1 Defining
Stoichiometry
pages 368–372
Practice Problems
pages 371–372
1. Interpret the following balanced chemical equa-
tions in terms of particles, moles, and mass.
Show that the law of conservation of mass is
observed.
a. N2(g) 3H2(g) → 2NH3(g)
1 molecule N2 3 molecules
H2 → 2 molecules NH3
1 mole N2 3 moles H2 → 2 moles NH3
Mass:
N2: (2 mol)(14.007 g/mol) 28.014 g
3H2: (6 mol)(1.008 g/mol) 6.048 g
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
2NH3: (2 mol)(14.007 g/mol) (6 mol)(1.008 g/mol) 34.062 g
c. 2Mg(s) O2(g) → 2MgO(s)
2 atoms Mg 1 molecule
O2 0 2 formula units MgO
2 moles Mg 1 mole O2 → 2 moles MgO
Mass:
2Mg: (2 mol)(24.305 g/mol) 48.610 g
O2: (2 mol)(15.999 g/mol) 31.998 g
2MgO: (2 mol)(24.305 g/mol) (2 mol)(15.999 g/mol) 80.608 g
48.610 g Mg 31.998 g O2 → 80.608 g MgO
80.608 g reactants 80.608 g products
2. Challenge For each of the following, balance
the chemical equation; interpret the equation
in terms of particles, moles, and mass; and
show that the law of conservation of mass is
observed.
a. ___Na(s) ___H2O(l) → ___NaOH(aq) ___H2(g)
28.014 g N2 6.048 g H2 0 34.062 g NH3
2Na(s) 2H2O(l) → 2NaOH(aq) 1H2(g)
34.062 g reactants 34.062 g products
2 atoms Na 2 molecules H2O →
2 formula units NaOH 1 molecule H2
b. HCl(aq) KOH(aq) → KCl(aq) H2O(l)
1 molecule HCl 1 formula unit KOH →
1 formula unit KCl 1 molecule H2O
1 mole HCl 1 mole KOH →
1 mole KCl 1 mole H2O
Mass:
HCl: (1 mol)(1.008 g/mol) (1 mol)(35.453 g/mol) 36.461 g
KOH: (1 mol)(39.098 g/mol) (1 mol)(15.999 g/mol) (1 mol)(1.008 g/mol)
56.105 g
2 mol Na 2 mol H2O →
2 mol NaOH 1 mol H2
Mass:
2Na: (2 mol)(22.990 g/mol) 45.980 g
2H2O: (4 mol)(1.008 g/mol) (2 mol)(15.999 g/mol) 36.030 g
2NaOH: (2 mol)(22.990 g/mol) (2 mol)(15.999 g/mol) (2 mol)(1.008 g/mol)
79.994 g
H2: (2 mol)(1.008 g/mol) 2.016 g
KCl: (1 mol)(39.098 g/mol) (1 mol)(35.453 g/mol) 74.551 g
45.980 g Na 36.030 g H2O →
79.994 g NaOH 2.016 g H2
H2O: (2 mol)(1.008 g/mol) (1 mol)( 15.999 g/mol) 18.015 g
82.01 g reactants 82.01 g products
36.461 g HCl 56.105 g KOH →
74.551 g KCl 18.015 g H2O
92.566 g reactants 92.566 g products
Solutions Manual
Chemistry: Matter and Change • Chapter 11
209
11
SOLUTIONS MANUAL
b. ___Zn(s) ___HNO3(aq) →
___Zn(NO3)2(aq) ___N2O(g) ___H2O(l)
b. 3Fe(s) 4H2O(l) → Fe3O4(s) 4H2(g)
3 mol Fe
3 mol Fe
3 mol Fe __
__
_
4 mol H2O
4Zn(s) 10HNO3(aq) →
4Zn(NO3)2(aq) 1N2O(g) 5H2O(l)
4 mol H2
1 mol Fe O
4 mol H
4 mol H O _
__
__
2
4 atoms Zn 10 molecules HNO3 →
4 formula units Zn(NO3)2 1 molecule N2O 5 molecules H2O
4 mol Zn 10 mol HNO3 →
4 mol Zn(NO3)2 1 mol N2O 5 mol H2O
Mass:
1 mol Fe3O4
3
2
3 mol Fe
3 mol Fe
4
3 mol Fe
1 mol Fe O
1 mol Fe O
4 mol H O
__
__
__
3
4
4 mol H2
3
4
2
4 mol H2O
4 mol H2
4 mol H O
4 mol H
4 mol H
__
__
__
2
2
1 mol Fe3O4
2
1 mol Fe3O4
4 mol H2O
c. 2HgO(s) → 2Hg(l) O2(g)
4Zn: (4 mol)(65.39 g/mol) 261.56 g
1 mol O
1 mol O
2 mol HgO _
__
__
10HNO3: (10 mol)(1.008 g/mol) (10 mol)(14.007 g/mol) (30 mol)(15.999 g/mol) 630.12 g
2
2 mol Hg
2
2 mol Hg
2 mol HgO
2 mol HgO
2 mol Hg __
2 mol Hg
__
_
4Zn(NO3)2: (4 mol)(65.39 g/mol) (8 mol)(14.007 g/mol) (24 mol)(15.999 g/mol) 757.592 g
2 mol HgO
1 mol O2
1 mol O2
4. Challenge Balance the following equations,
and determine the possible mole ratios.
a. ZnO(s) HCl(aq) → ZnCl2(aq) H2O(l)
N2O: (2 mol)(14.007 g/mol) (1 mol)(15.999 g/mol) 44.013 g
ZnO(s) 2HCl(aq) → ZnCl2(aq) H2O(l)
5H2O: (10 mol)(1.008 g/mol) (5 mol)(15.999 g/mol) 90.075 g
1 mol ZnO __
1 mol ZnO
1 mol ZnO
__
__
2 mol HCl
1 mol ZnCl2
1 mol H2O
261.56 g Zn 630.12 g HNO3 →
757.592 g Zn(NO3)2 44.013 g N2O 90.075 g H2O
2 mol HCl
2 mol HCl
2 mol HCl
__
__
__
891.68 g reactants 891.68 g products
1 mol ZnCl
1 mol ZnCl
1 mol ZnCl
__
__
__
3. Determine all possible mole ratios for the
following balanced chemical equations.
a. 4Al(s) 3O2(g) → 2Al2O3(s)
2 mol Al O
3 mol O
4 mol Al __
_
__
2
2
3 mol O2
2 mol Al2O3
3
4 mol Al
2 mol Al O
3 mol O
4 mol Al
_
__
__
2
4 mol Al
2
3 mol O2
3
2 mol Al2O3
1 mol ZnO
1 mol ZnCl2
2
1 mol ZnO
1 mol H2O
2
2 mol HCl
2
1 mol H2O
1 mol H O __
1 mol H O
1 mol H O __
__
2
2
1 mol ZnO
2
2 mol HCl
1 mol ZnCl2
b. butane (C4H10) oxygen →
carbon dioxide water
2C4H10(g) 13O2(g) 0 8CO2(g) 10H2O(l)
2 mol C H
2 mol C H
2 mol C H
__
__
__
4 10
13 mol O2
4 10
8 mol CO2
4 10
10 mol H2O
8 mol CO
10 mol H O
13 mol O
__
__
__
2
2 mol C4H10
2
2 mol C4H10
2
2 mol C4H10
10 mol H O _
8 mol CO
10 mol H O __
__
2
2
13 mol O2
2
8 mol CO2
13 mol O2
8 mol CO
13 mol O
13 mol O
__
__
_
2
10 mol H2O
210
Chemistry: Matter and Change • Chapter 11
2
10 mol H2O
2
8 mol CO2
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CHAPTER
CHAPTER
11
Section 11.1 Assessment
page 372
5. Compare the mass of the reactants and the
mass of the products in a chemical reaction, and
explain how these masses are related.
SOLUTIONS MANUAL
10. Model Write the mole ratios for the reaction
of hydrogen gas and oxygen gas, 2H2(g) O2(g) → 2H2O. Make a sketch of six hydrogen
molecules reacting with the correct number of
oxygen molecules. Show the water molecules
produced.
The coefficients in the balanced equation indicate
the molar relationship between each pair of
reactants and products. The masses of reactants
and products are equal.
6. State how many mole ratios can be written for
a chemical reaction involving three substances.
n 3, thus (n)(n 1) (3)(2) 6 mole ratios
7. Categorize the ways in which a balanced
chemical equation can be interpreted.
particles (atoms, molecules, formula units), moles,
and mass
8. Apply The general form of a chemical reac-
tion is xA yB → zAB. In the equation, A and
B are elements and x, y, and z are coefficients.
State the mole ratios for this reaction.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
xA/yB and xA/zAB
6H2 3O2
6H2O
2H2/O2 and 2H2/2H2O
O2/2H2 and O2/2H2O
2H2O/2H2 and 2H2O/O2
Sketches should show six hydrogen atoms
combining with three oxygen atoms to form six
water molecules.
yB/xA and yB/zAB
zAB/xA and zAB/yB
9. Apply Hydrogen peroxide (H2O2) decomposes
to produce water and oxygen. Write a balanced
chemical equation for this reaction, and determine the possible mole ratios.
2H2O2 → 2H2O O2
2 mol H2O2/2 mol H2O, 2 mol H2O2/1 mol O2,
2 mol H2O/2 mol H2O2, 2 mol H2O/1 mol O2,
1 mol O2/2 mol H2O2, 1 mol O2/2 mol H2O
Section 11.2 Stoichiometric
Calculations
pages 373–378
Practice Problems
pages 375–377
11. Methane and sulfur react to produce carbon
disulfide (CS2), a liquid often used in the
production of cellophane.
___CH4(g) ___S8(s) → ___CS2(l) ___H2S(g)
a. Balance the equation.
2CH4(g) S8(s) → 2CS2(l) 4H2S(g)
b. Calculate the moles of CS2 produced when
1.50 moles of S8 is used.
1.50 mol S8 Solutions Manual
2 mol CS
_
3.00 mol CS
2
1 mol S8
Chemistry: Matter and Change • Chapter 11
2
211
11
SOLUTIONS MANUAL
c. How many moles of H2S are produced?
1.50 mol S8 4 mol H S
_
6.00 mol H S
2
2
1 mol S8
12. Challenge Sulfuric acid (H2SO4) is formed
when sulfur dioxide (SO2) reacts with oxygen
and water.
a. Write the balanced chemical equation for the
reaction.
2SO2(g) O2(g) 2H2O(l) → 2H2SO4(aq)
b. How many moles of H2SO4 are produced
from 12.5 moles of SO2?
12.5 mol SO2 __
elements sodium and chlorine by means of
electrical energy, as shown below. How much
chlorine gas, in grams, is obtained from the
process?
Na
2
2
1 mol Cl2
2
1 mol C
__
1.25 mol C
1 mol TiO2
Step 2: Make mole → mass conversion.
1.25 mol C 12.011 g C
__
15.0 g C
1 mol C
c. What is the mass of all of the products
formed by reaction with 1.25 mol of TiO2?
Cl2 ? g
Calculate the mass to TiO2 used.
Step 1: Balance the chemical equation.
1.25 mol TiO2 2NaCl(s) → 2Na(s) Cl2(g)
79.865 g TiO
__
99.8 g TiO
2
1 mol TiO2
2
Calculate the total mass of the reactants.
Step 2: Make mole → mole conversion.
1 mol Cl
__
1.25 mol Cl
99.8 g 177 g 15.0 g 292 g
2
2
2 mol NaCl
Step 3: Make mole → mass conversion.
212
2
1 mol TiO2
70.90 g Cl
__
177 g Cl
1.25 mol TiO2 13. Sodium chloride is decomposed into the
1.25 mol Cl2
2 mol Cl
__
2.50 mol Cl
Step 1: Make mole → mole conversion.
2 mol SO2
6.25 mol O2 needed
2.50 mol NaCl 1.25 mol TiO2 1.25 mol of TiO2?
2
2.50 mol
Step 1: Make mole → mole conversion.
b. What mass of C is needed to react with
1 mol O
_
NaCl
TiO2(s) C(s) 2Cl2(g) → TiCl4(s) CO2(g)
a. What mass of Cl2 gas is needed to react with
1.25 mol of TiO2?
2.50 mol Cl2 c. How many moles of O2 are needed?
Electric
energy
in many alloys because it is extremely strong
and lightweight. Titanium tetrachloride (TiCl4)
is extracted from titanium oxide (TiO2) using
chlorine and coke (carbon).
Step 2: Make mole → mass conversion.
2 mol H2SO4
2 mol SO2
12.5 mol H2SO4 produced
12.5 mol SO2
14. Challenge Titanium is a transition metal used
70.9 g Cl
_ 88.6 g Cl
Because mass is conserved, the mass of the
products must equal the mass of reactants.
mass of products 292 g
2
1 mol Cl2
2
Chemistry: Matter and Change • Chapter 11
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CHAPTER
CHAPTER
11
SOLUTIONS MANUAL
Step 4: Make mole → mass conversion.
15. One of the reactions used to inflate
automobile air bags involves sodium azide
(NaN3): 2NaN3(s) → 2Na(s) 3N2(g).
Determine the mass of N2 produced from
the decomposition of NaN3 shown below.
0.0390 mol H2SO4 3.83 g H2SO4
98.09 g H SO
__
2
4
1 mol H2SO4
Section 11.2 Assessment
N2 gas
page 378
17. Explain why a balanced chemical equation is
needed to solve a stoichiometric problem.
The coefficients in the balanced equation indicate
the molar relationship between each pair of
reactants and products.
100.0 g NaN3 → ? g N2(g)
2NaN3(s) → 2Na(s) 3N2(g)
Step 1: Make mass → mole conversion.
100.0 g NaN3 18. List the four steps used in solving stoichio-
__ 1.538 mol NaN
1 mol NaN3
3
65.02 g NaN3
Step 2: Make mole → mole conversion.
1.538 mol NaN3 3 mol N
__
2.307 mol N
2
2
2 mol NaN3
Step 3: Make mole → mass conversion.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
2.307 mol N2 28.02 g N
__
64.64 g N
2
16. Challenge In the formation of acid rain, sulfur
dioxide (SO2) reacts with oxygen and water
in the air to form sulfuric acid (H2SO4). Write
the balanced chemical equation for the reaction. If 2.50 g of SO2 reacts with excess oxygen
and water, how much H2SO4, in grams, is
produced?
Step 1: Balance the chemical equation.
2SO2(g) O2(g) 2H2O(l) 0 2H2SO4(aq)
Step 2: Make mass → mole conversion.
2.50 g SO2 1. Balance the equation.
2. Convert the mass of the known substance to
moles of known substance.
3. Use the mole ratio to convert from moles of
the known to moles of the unknown.
4. Convert moles of unknown to mass of the
unknown.
2
1 mol N2
metric problems.
1 mol SO
__
0.0390 mol SO
2
19. Describe how a mole ratio is correctly
expressed when it is used to solve a stoichiometric problem.
moles of unknown/moles of known
20. Apply How can you determine the mass of
liquid bromine (Br2) needed to react completely
with a given mass of magnesium?
Write a balanced equation. Convert the given
mass of magnesium to moles. Use the mole ratio
from the balanced equation to convert moles of
magnesium to moles of bromine. Convert from
moles of bromine to mass of bromine.
2
64.07 g SO2
Step 3: Make mole → mole conversion.
0.0390 mol SO2 2 mol H SO
__
2
4
2 mol SO2
0.0390 mol H2SO4
Solutions Manual
Chemistry: Matter and Change • Chapter 11
213
11
SOLUTIONS MANUAL
21. Calculate Hydrogen reacts with excess
nitrogen as follows:
0.6261 mol Fe O
1 mol Fe O
__
compared to __
2
If 2.70 g of H2 reacts, how many grams of NH3
is formed?
( __ )
1.34 mol H2
1 mol
1.34 mol H2
2.016 g H2
(_)
( __ )
2 mol NH3
3 mol H2
(0.893 mol NH3)
3
2
4.350 mol Na
N2(g) 3H2(g) → 2NH3(g)
(2.70 g H2)
Make mole ratio comparison.
0.893 mol NH3
17.031 g NH3
1 mol NH3
0.1439 compared to 0.1667
The actual ratio is less than the needed ratio,
so iron(III) oxide is the limiting reactant.
b. reactant in excess
Sodium is the excess reactant.
c. mass of solid iron produced
15.2 g NH3
22. Design a concept map for the following
Make mole → mole conversion.
0.6261 mol Fe2O3 2 mol Fe
__
1 mol Fe2O3
reaction.
1.252 mol Fe
CaCO3(s) 2HCl(aq) → CaCl2(aq) H2O(l) CO2(g)
Make mole → mass conversion.
The concept map should explain how to determine the mass of CaCl2 produced from a given
mass of HCl.
Concept maps will vary, but all should show the
use of these conversion factors: the inverse of
molar mass, the mole ratio, the molar mass.
Section 11.3 Limiting Reactants
pages 379–384
1.252 mol Fe 1 mol Fe
the reaction is complete
Make mole → mole conversion.
0.6261 mol Fe2O3 6 mol Na
__
1 mol Fe2O3
3.757 mol Na needed
Make mole → mass conversion.
22.9 g Na
_
1 mol Na
86.37 g Na needed
page 383
23. The reaction between solid sodium and iron(III)
oxide is one in a series of reactions that inflates
an automobile airbag: 6Na(s) Fe2O3(s) →
3Na2O(s) 2Fe(s). If 100.0 g of Na and 100.0 g
of Fe2O3 are used in this reaction, determine the
following.
a. limiting reactant
Make mass → mole conversion.
1 mol Na
__
4.350 mol Na
22.99 g Na
100.0 g Fe2O3 55.85 g Fe
_
69.92 g Fe
d. mass of excess reactant that remains after
3.757 mol Na Practice Problems
100.0 g Na 3
6 mol Na
100.0 g Na given 86.37 g Na needed
13.6 g Na in excess
24. Challenge Photosynthesis reactions in green
plants use carbon dioxide and water to produce
glucose (C6H12O6) and oxygen. A plant has
88.0 g of carbon dioxide and 64.0 g of water
available for photosynthesis.
a. Write the balanced chemical equation for the
reaction.
6CO2(g) 6H2O(l) 0 C6H12O6(aq) 6O2(g)
1 mol Fe O
__
2
3
159.7 g Fe2O3
0.6261 mol Fe2O3
214
Chemistry: Matter and Change • Chapter 11
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CHAPTER
11
CHAPTER
SOLUTIONS MANUAL
Section 11.3 Assessment
b. Determine the limiting reactant.
page 384
Make mass → mole conversion.
88.0 g CO2
1 mol CO
__ 2.00 mol CO
2
2
44.01 g CO2
1 mol H O
64.0 g H O __ 3.55 mol H O
25. Describe the reason why a reaction between
two elements comes to an end.
one of the reactants is used up
2
2
2
18.02 g H2O
each reaction.
a. Wood burns in a campfire.
Make mole ratio comparison.
2.00 mol CO
6 mol CO
__
compared to __ ;
2
2
3.55 mol H2O
6 mol H2O
26. Identify the limiting and the excess reactant in
The wood limits. Oxygen is in excess. The fire
will burn only while wood is present.
0.563 compared to 1.00
The actual ratio is less than the needed ratio,
so carbon dioxide is the limiting reactant.
b. Airborne sulfur reacts with the silver plating
on a teapot to produce tarnish (silver sulfide).
Silver is the limiting reactant. Sulfur is in
excess. When a layer of tarnish covers the
silver surface, it prevents the sulfur in the air
from reacting.
c. Determine the excess reactant.
Water is the excess reactant.
d. Determine the mass in excess.
c. Baking powder in batter decomposes to
Make mole → mole conversion.
2.00 mol CO2 __ 2.00 mol H O
6 mol H2O
2
6 mol CO2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Make mole → mass conversion.
2.00 mol H2O 18.02 g H O
__
64.0 g H2O given 36.0 g H2O needed
28.0 g H2O in excess
e. Determine the mass of glucose produced.
Make mole → mole conversion.
1 mol C H O
__
6 12
6
6 mol CO2
0.333 mol C6H12O6
Make mole 0 mass conversion.
0.333 mol C6H12O6 60.0 g C6H12O6
A decomposition reaction usually has only
one reactant. The reaction is limited by the
amount of baking powder present.
27. Analyze Tetraphosphorous trisulphide (P4S3)
2
1 mol H2O
36.0 g H2O needed
2.00 mol CO2 produce carbon dioxide.
__
is used in the match heads of some matches. It
is produced in the reaction 8P4 3S8 → 8P4S3.
Determine which of the following statements
are incorrect, and rewrite the incorrect statements to make them correct.
a. 4 mol P4 reacts with 1.5 mol S8 to form
4 mol P4S3.
correct
b. Sulfur is the limiting reactant when 4 mol P4
and 4 mol S8 react.
Phosphorus is the limiting reactant.
180.24 g C6H12O6
1 mol C6H12O6
c. 6 mol P4 reacts with 6 mol S8, forming
1320 g P4S3.
correct
Solutions Manual
Chemistry: Matter and Change • Chapter 11
215
11
SOLUTIONS MANUAL
Section 11.4 Percent Yield
b. Determine the percent yield if 515.6 g of
pages 385–388
product is recovered.
Practice Problems
% yield 28. Aluminum hydroxide (Al(OH)3) is often present
in antacids to neutralize stomach acid (HCl).
The reaction occurs as follows: Al(OH)3(s) 3HCl(aq) → AlCl3(aq) 3H2O(l). If 14.0 g of
Al(OH)3 is present in an antacid tablet, determine the theoretical yield of AlCl3 produced
when the tablet reacts with HCl.
1 mol Al(OH)
__
determine the theoretical yield of silver.
Make mass → mole conversion.
Make mole → mole conversion.
20.0 g Cu __
1 mol AlCl3
1 mol Al(OH)3
63.55 g Cu
0.315 mol Cu 133.3 g AlCl
__
23.9 g AlCl
2 mol Ag
_
0.630 mol Ag
1 mol Cu
Make mole → mass conversion.
3
1 mol AlCl3
1 mol Cu
__
0.315 mol Cu
Make mole → mole conversion.
Make mole → mass conversion.
0.179 mol AlCl3 silver nitrate solution (AgNO3), silver crystals
and copper(II) nitrate (Cu(NO3)2) solution
form.
a. Write the balanced chemical equation for the
reaction.
b. If a 20.0g sample of copper is used,
3
78.0 g Al(OH)3
0.179 mol Al(OH)3
0.179 mol AlCl3
30. Challenge When copper wire is placed into a
Cu(s) 2AgNO3(aq) → 2Ag(s) Cu(NO3)2(aq)
Make mass → mole conversion.
0.179 mol Al(OH)3 2
610.3 g ZnI2
84.48% yield of ZnI2
page 387
14.0 g Al(OH)3 515.6 g ZnI
__
100
3
0.630 mol Ag 23.9 g of AlCl3 is the theoretical yield.
29. Zinc reacts with iodine in a synthesis reaction:
Zn I2 0 ZnI2.
a. Determine the theoretical yield if 1.912 mol
of zinc is used.
1 mol Ag
68.0 g of Ag is the theoretical yield.
c. If 60.0 g of silver is recovered from the
reaction, determine the percent yield of the
reaction.
% yield Write the balanced chemical equation.
107.9 g Ag
__
68.0 g Ag
60.0 g Ag
_
100
68.0 g Ag
88.2% yield of Ag
Zn(s) I2(s) → ZnI2(s)
Make mole → mole conversion.
1.912 mol Zn Section 11.4 Assessment
1 mol ZnI
_
1.912 mol ZnI
2
1 mol Zn
2
31. Identify which type of yield—theoretical
Make mole → mass conversion.
1.912 mol ZnI2 __ 610.3 g ZnI
319.2 g ZnI2
1 mol ZnI2
2
610.3 g of ZnI2 is the theoretical yield.
216
page 388
Chemistry: Matter and Change • Chapter 11
yield, actual yield, or percent yield—is a
measure of the efficiency of a chemical
reaction.
percent yield
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CHAPTER
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CHAPTER
SOLUTIONS MANUAL
32. List several reasons why the actual yield from
a chemical reaction is not usually equal to the
theoretical yield.
Not all reactions go to completion. Some of the
reactants or products stick to the surface of the
container and are not massed or transferred.
Other unexpected products form from competing
reactions.
33. Explain how percent yield is calculated.
divide the actual yield by the theoretical yield and
multiply the quotient by 100
34. Apply In an experiment, you combine 83.77 g
of iron with an excess of sulfur and then heat
the mixture to obtain iron(III) sulfide.
What is the theoretical yield, in grams, of
iron(III) sulfide?
( __ )
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
1.500 mol Fe
1 mol Fe
1.500 mol Fe
55.845 g Fe
( __ ) ( __ )
1 mol Fe2S3
207.885 g Fe2S3
2 mol Fe
1 mol Fe2S3
155.9 g Fe2S3
pages 392–397
Section 11.1
Mastering Concepts
36. Why must a chemical equation be balanced
before you can determine mole ratios?
Mole ratios are determined by the coefficients
in a balanced equation. If the equation is not
balanced, the relationship between reactants and
products cannot be determined.
37. What relationships can be determined from a
balanced chemical equation?
The relationships among particles, moles, and
mass for all reactants and products.
2Fe(s) 3S(s) → Fe2S3(s)
(83.77 g Fe)
Chapter 11 Assessment
38. Explain why mole ratios are central to
stoichiometric calculations.
Mole ratios allow for the conversion from moles of
one substance in a balanced chemical equation to
moles of another substance in the same equation.
39. What is the mole ratio that can convert from
moles of A to moles of B?
35. Calculate the percent yield of the reaction of
2Mg(s) O2(g) → 2MgO(s)
40. Why are coefficients used in mole ratios instead
of subscripts?
Reaction Data
Mass of empty crucible
35.67 g
Mass of crucible and Mg
38.06 g
Mass of crucible and MgO (after
heating)
39.15 g
mass of Mg 38.06 g 35.67 g 2.39 g
mass of MgO 39.15 g 33.67 g 3.48 g
Theoretical yield: 2.39 g Mg moles B
_
moles A
magnesium with excess oxygen:
1 mol Mg
__
24.31 g Mg
The coefficients in the balanced chemical
equation show the numbers of representative
particles involved in a reaction. Subscripts give
the numbers of different kinds of atoms within a
molecule or formula unit.
41. Explain how the conservation of mass allows
you to interpret a balanced chemical equation in
terms of mass.
The mass of the reactants will always equal the
mass of the products.
0.0983 mol Mg
0.983 mol Mg 40.31 g MgO
2 mol MgO
__
__
2 mol Mg
1 mol MgO
3.96 g MgO
% yield 3.48 g MgO
__
100
3.96 g MgO
87.9% yield of MgO
Solutions Manual
Chemistry: Matter and Change • Chapter 11
217
11
CHAPTER
SOLUTIONS MANUAL
42. When heated by a flame, ammonium dichro-
mate decomposes, producing nitrogen gas, solid
chromium(III) oxide, and water vapor.
Mastering Problems
44. Interpret the following equation in terms of
particles, moles, and mass.
(NH4)2Cr2O7 → N2 Cr2O3 4H2O
4Al(s) 3O2(g) → 2Al2O3(s)
Write the mole ratios for this reaction that relate
ammonium chromate to the products.
Particles: 4 atoms Al 3 molecules O2 →
2 formula units Al2O3
1 mol (NH ) Cr O
1 mol (NH ) Cr O
__
__
Moles: 4 mol Al 3 mol O2 → 2 mol Al2O3;
4 2
2
7
4 2
1 mol N2
2
7
1 mol Cr2O3
1 mol (NH ) Cr O
__
4 2
2
Mass:
4Al 4 mol Al(26.982 g/mol) 107.93 g
7
4 mol H2O
__ __
1 mol N2
1 mol Cr2O3
1 mol (NH4)2Cr2O7
1 mol (NH4)2Cr2O7
4 mol H O
__
3O2 6 mol O(15.999 g/mol) 95.99 g
2Al2O3 4 mol Al(26.982 g/mol) 6 mol O(15.999 g/mol) 203.92 g
107.93 g Al 95.99 g O2 → 203.92 g Al2O3
2
1 mol (NH4)2Cr2O7
45. Smelting When tin(IV) oxide is heated with
carbon in a process called smelting, the element tin
can be extracted.
SnO2(s) 2C(s) → Sn(l) 2CO(g)
representing Element M and circles representing Element N. Write a balanced equation
to represent the picture shown, using smallest
whole-number ratios. Write mole ratios for this
equation.
Particles: 1 formula unit SnO2 2 atoms C → 1 atom Sn 2 molecules CO
2M2N → M4 N2
Mass:
2 mol M N __
2 mol M N
__
,
,
2
1 mol M4
2
1 mol N2
1 mol M
1 mol M __
_
,
,
4
4
Moles: 1 mol SnO2 2 mol
C → 1 mol Sn 2 mol CO
SnO2: 1 mol(118.710 g/mol) 2 mol (15.999 g/mol) 150.71 g
2C: 2 mol(12.011 g/mol) 24.02 g
1 mol N2 2 mol M2N
Sn: 1 mol(118.710 g/mol) 118.71 g
1 mol N
1 mol N
__
,_
2CO: 2 mol(12.011 g/mol) 2 mol
(15.999 g/mol) 56.02 g
2
2
2 mol M2N 1 mol M4
218
Interpret the chemical equation in terms of
particles, moles, and mass.
Chemistry: Matter and Change • Chapter 11
150.71 g SnO2 24.02 g C →
118.71 g Sn 56.02 g CO
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
43. Figure 11.10 depicts an equation with squares
CHAPTER
11
46. When solid copper is added to nitric acid,
48. When aluminum is mixed with iron(III) oxide,
copper(II) nitrate, nitrogen dioxide, and water
are produced. Write the balanced chemical
equation for the reaction. List six mole ratios
for the reaction.
iron metal and aluminum oxide are produced
along with a large quantity of heat. What mole
ratio would you use to determine moles of Fe if
moles of Fe2O3 are known?
Cu(s) 4HNO3(aq) → Cu(NO3)2(aq) 2NO2(g) 2H2O(l); answers should include any six of the
following ratios: 1 mol Cu/4 mol HNO3,
1 mol Cu/1 mol Cu(NO3)2, 1 mol Cu/2 mol NO2,
1 mol Cu/2 mol H2O, 4 mol HNO3/1 mol Cu, 4 mol
HNO3/1 mol Cu(NO3)2, 4 mol HNO3/2 mol NO2,
4 mol HNO3/2 mol H2O, 1 mol Cu(NO3)2/1 mol
Cu, 1 mol Cu(NO3)2/4 mol HNO3, 1 mol
Cu(NO3)2/2 mol NO2, 1 mol Cu(NO3)2/2 mol H2O,
2 mol NO2/1 mol Cu, 2 mol NO2/4 mol
HNO3, 2 mol NO2/1 mol Cu(NO3)2, 2 mol
NO2/2 mol H2O, 2 mol H2O/1 mol Cu, 2 mol
H2O/4 mol HNO3, 2 mol H2O/1 mol Cu(NO3)2,
2 mol H2O/2 mol NO2
Fe2O3(s) 2Al(s) → 2Fe(s) Al2O3(s) heat
47. When hydrochloric acid solution reacts with
lead(II) nitrate solution, lead(II) chloride precipitates and a solution of nitric acid is produced.
a. Write the balanced chemical equation for the
reaction.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
SOLUTIONS MANUAL
2HCl(aq) Pb(NO3)2(aq) →
PbCl2(s) 2HNO3(aq)
b. Interpret the equation in terms of molecules
and formula units, moles, and mass.
Particles: 2 molecules HCl 1 formula unit
Pb(NO3)2 → 1 formula unit PbCl2 2 molecules HNO3
Moles: 2 mol HCl 1 mol Pb(NO3)2 →
1 mol PbCl2 2 mol HNO3;
2 mol Fe
__
1 mol Fe2O3
49. Solid silicon dioxide, often called silica, reacts
with hydrofluoric acid (HF) solution to produce
the gas silicon tetrafluoride and water.
a. Write the balanced chemical equation for the
reaction.
SiO2(s) 4HF(aq) → SiF4(g) 2H2O(l)
b. List three mole ratios, and explain how
you would use them in stoichiometric
calculations.
Students may write any of 12 ratios. Examples
might be the following. 4 mol HF/1 mol
SiO2; used to find the amount of HF that will
react with a known amount of SiO2; 1 mol
SiF4/1 mol SiO2; used to find the amount of
SiF4 that can be formed from a known amount
of SiO2. 2 mol H2O/1 mol SiF4; used to find the
amount of H2O that will be produced with
the SiF4
50. Chrome The most important commercial ore
of chromium is chromite (FeCr2O4). One of the
steps in the process used to extract chromium
from the ore is the reaction of chromite with
coke (carbon) to produce ferrochrome (FeCr2).
Mass:
2C(s) FeCr2O4(s) → FeCr2(s) 2CO2(g)
2HCl: 2 mol(1.008 g/mol) 2 mol (35.453 g/mol) 72.9 g
What mole ratio would you use to convert from
moles of chromite to moles of ferrochrome?
Pb(NO3)2: 1 mol(207.2 g/mol) 2 mol (14.007 g/mol) 6 mol(15.999 g/mol)
331.2 g
1 mol FeCr
__
2
1 mol FeCr2O4
PbCl2: 1 mol(207.2 g/mol) 2 mol
(35.453 g/mol) 278.1 g
2HNO3: 2 mol(1.008 g/mol) 2 mol
(14.007 g/mol) 6 mol(15.999 g/mol) 126.0 g
72.9 g HCl 331.2 g Pb(NO3)2 →
278.1 g PbCl2 126.0 g HNO3
Solutions Manual
Chemistry: Matter and Change • Chapter 11
219
11
SOLUTIONS MANUAL
51. Air Pollution The pollutant SO2 is removed
from the air in a reaction that also involves
calcium carbonate and oxygen. The products
of this reaction are calcium sulfate and carbon
dioxide. Determine the mole ratio you would
use to convert moles of SO2 to moles of CaSO4.
2SO2 2CaCO3 O2 → 2CaSO4 2CO2
2 mol CaSO
__
4
Section 11.2
Mastering Concepts
54. What is the first step in all stoichiometric
calculations?
Write a balanced chemical equation for the
reaction.
55. What information does a balanced equation
2 mol SO2
provide?
52. Two substances, W and X, react to form the
products Y and Z. Table 11.2 shows the moles
of the reactants and products involved when
the reaction was carried out. Use the data to
determine the coefficients that will balance the
equation W X → Y Z.
Reaction Data
Moles of Reactants
Moles of Products
W
X
Y
Z
0.90
0.30
0.60
1.20
Divide each molar quantity by 0.30 mol, the least
common denominator:
The balanced equation provides the relationship
between reactants and products, and the
coefficients in the equation are used to write
mole ratios relating reactants and products.
56. On what law is stoichiometry based, and how
do the calculations support this law?
Stoichiometry is based on the law of conservation
of mass. The calculations are used to determine
the mass of reactants and products. Once found,
the sum of reactants will equal the sum of
products, verifying the law of conservation of
mass.
57. How is molar mass used in some stoichiometric
calculations?
W: 0.90/0.30 3
Molar mass is a conversion factor for converting
moles of a given substance to mass or mass of a
given substance to moles.
X: 0.30/0.30 1
Y: 0.60/0.30 2
Z: 1.20/0.30 4
58. What information must you have in order to
calculate the mass of product formed in a
chemical reaction?
3W X 0 2Y 4Z
53. Antacids Magnesium hydroxide is an ingre-
dient in some antacids. Antacids react with
excess hydrochloric acid in the stomach to
relieve indigestion.
You must have the balanced chemical equation
and know the quantity of one substance in
the reaction other than the product you are to
determine.
___Mg(OH)2 ___HCl →
___ MgCl2 ___ H2O
a. Balance the reaction of Mg(OH)2 with HCl.
1Mg(OH)2 2HCl → 1MgCl2 2H2O
b. Write the mole ratio that would be used to
determine the number of moles of MgCl2
produced when HCl reacts with Mg(OH)2.
1 mol MgCl
__
2
1 mol Mg(OH)2
220
or
1 mol MgCl
__
2
2 mol HCl
Chemistry: Matter and Change • Chapter 11
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
CHAPTER
CHAPTER
11
SOLUTIONS MANUAL
61. Welding If 5.50 mol of calcium carbide (CaC2)
reacts with an excess of water, how many moles
of acetylene (C2H2), a gas used in welding, will be
produced?
+
CaC2(s) 2H2O(l) → Ca(OH)2(aq) C2H2(g)
59. Each box in Figure 11.11 represents the
contents of a flask. One flask contains hydrogen
sulfide, and the other contains oxygen. When
the contents of the flasks are mixed, a reaction
occurs and water vapor and sulfur are produced.
In the figure, the red circles represent oxygen,
the yellow circles represent sulfur, and blue
circles represent hydrogen.
a. Write the balanced chemical equation for the
reaction.
2H2S(g) O2(g) → 2H2O(g) 2S(s)
b. Using the same color code, sketch a
representation of the flask after the reaction
occurs.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Student sketches should show the formation
of six water molecules (H2O) and six sulfur
atoms (S).
Mastering Problems
60. Ethanol (C2H5OH ), also known as grain
alcohol, can be made from the fermentation
of sugar (C6H12O6). The unbalanced chemical
equation for the reaction is shown below.
___C6H12O6 → ___C2H5OH ___CO2
Balance the chemical equation and determine
the mass of C2H5OH produced from 750 g of
C6H12O6.
C6H12O6 → 2C2H5OH 2CO2
(
2 mol C H OH
__
8.4 mol C2H5OH
8.4 mol C2H5OH
Solutions Manual
(
2 5
1 mol C6H12O6
46.07 g
__
1 mol C2H5OH
62. Antacid Fizz When an antacid tablet dissolves
in water, the fizz is due to a reaction between
sodium hydrogen carbonate (NaHCO3), also
called sodium bicarbonate, and citric acid
(H3C6H5O7).
3NaHCO3(aq) H3C6H5O7(aq) →
3CO2(g) 3H2O(l) Na3C6H5O7(aq)
How many moles of Na3C6H5O7 can be
produced if one tablet containing 0.0119 mol of
NaHCO3 is dissolved?
0.0119 mol NaHCO3 (1 mol Na3C6H5O7/3 mol
NaHCO3) 0.00397 mol
63. Esterification The process in which an
organic acid and an alcohol that forms as ester
and water is known as esterification. Ethyl
butanoate (C3H7COOC2H5), an ester, is formed
when the alcohol ethanol (C2H5OH) and
butanoic acid (C3H7COOH) are heated in the
presence of sulfuric acid.
C2H5OH(l) C3H7COOH(l) →
C3H7COOC2H5(l) H2O(l)
Determine the mass of ethyl butanoate produced
if 4.50 mol of ethanol is used.
1 mol C H COOC H
__
1 mol C H OH
116.18
g
C
H
COOC
H
___
3 7
4.50 mol C2H5OH 2 5
2 5
750 g C6H12O6(180.16 g/mol) 4.2 mol C6H12O6
4.2 mol C6H12O6
The mole ratio of CaC2:C2H2 is 1:1; thus, 5.50 mol
C2H2 will be produced from 5.50 mol CaC2.
)
)
3 7
2 5
1 mol C3H7COOC2H5
523 g C3H7COOC2H5
390 g C2H5OH
Chemistry: Matter and Change • Chapter 11
221
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SOLUTIONS MANUAL
64. Greenhouse Gas Carbon dioxide is a green-
house gas that is linked to global warming.
It is released into the atmosphere through the
combustion of octane (C8H18) in gasoline. Write
the balanced chemical equation for the combustion of octane and calculate the mass of octane
needed to release 5.00 mol of CO2.
2C8H18(l) 25O2(g) 0 16CO2(g) 18H2O(l)
5.00 mol CO2 0.625 mol C8H18
__ 0.625 mol C H
2 mol C8H18
produced by a reaction between methane and
chlorine.
CH4(g) 3Cl2(g) 0 CHCl3(g) 3HCl(g)
How much CH4 in grams is needed to produce
50.0 grams of CHCl3?
50.0 g CHCl3 8 18
16 mol CO2
114.28 g C H
__ 71.4 g C H
8 18
1 mol C8H18
solution of lead(II) nitrate to produce a yellow
precipitate of lead(II) chromate and a solution
of potassium nitrate.
a. Write the balanced chemical equation.
K2CrO4(aq) Pb(NO3)2(aq) 0
PbCrO4(s) 2KNO3(aq)
chromate, determine the mass of lead
chromate formed.
1 mol PbCrO
__
4
1 mol CHCl3
4
4
1 mol CH4
68. Oxygen Production The Russian Space
Agency uses potassium superoxide (KO2) for
the chemical oxygen generators in their space
suits.
4KO2 2H2O 4CO2 0 4 KHCO3 3O2
Complete Table 11.3.
Mass
KO2
Mass
H2O
Mass
CO2
Mass
KHCO3
Mass
O2
1100 g
140 g
7.0 102 g
1600 g
380 g
380 g O2 (1 mol O2/32.00 g O2)
(4 mol KO2/3 mol O2) (71.1 g KO2/1 mol KO2)
1100 g
4
1 mol K2CrO4
323.2 g PbCrO
__
80.8 g PbCrO
4
4
1 mol PbCrO4
3
119.37 g CHCl3
Oxygen Generation Reaction Data
b. Starting with 0.250 mol of potassium
0.250 mol K2CrO4 1 mol CHCl
1 mol CH
__
__
16.04 g CH
__
6.72 g CH
8 18
65. A solution of potassium chromate reacts with a
67. Chloroform (CHCl3), an important solvent, is
66. Rocket Fuel The exothermic reaction between
liquid hydrazine (N2H2) and liquid hydrogen
peroxide (H2O2) is used to fuel rockets. The
products of this reaction are nitrogen gas and
water.
a. Write the balanced chemical equation.
N2H2(l) H2O2(l) 0 N2(g) 2H2O(g)
380 g O2 (1 mol O2/32.00 g O2)
(2 mol H2O/3 mol O2)
(18.02 g H2O/1 mol H2O) 140 g
380 g O2 (1 mol O2/32.00 g O2)
(4 mol CO2/3 mol O2)
(44.01 g CO2/1 mol CO2) 7.0 102 g
380 g O2 (1 mol O2/32.00 g O2)
(4 mol KHCO3/3 mol O2)
(100.12 g KHCO3/1 mol KHCO3) 1600 g
b. How much hydrazine, in grams, is needed to
produce 10.0 mol of nitrogen gas?
10.0 mol N2 1 mol N H
30.03 g N H
__
__
2 2
1 mol N2
3.00 102 g N2H2
222
2 2
1 mol N2H2
Chemistry: Matter and Change • Chapter 11
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CHAPTER
CHAPTER
11
SOLUTIONS MANUAL
69. Gasohol is a mixture of ethanol and gasoline.
was extracted is 150.0 g, what percentage of
the ore is gold?
C2H5OH(l) O2(g) 0 CO2(g) H2O(g)
150.0 g ore
C2H5OH(l) 3O2(g) 0 2CO2(g) 3H2O(l)
1 mol C H OH
100.0 g C H OH __
2 5
2 5
46.08 g C2H5OH
2.170 mol C2H5OH
2.170 mol C2H5OH 4.340 mol CO2
4.340 mol CO2
2 mol CO
__
2
1 mol C2H5OH
50.2 g Au
__
100 33.5% gold in ore
72. Film Photographic film contains silver bromide
in gelatin. Once exposed, some of the silver
bromide decomposes, producing fine grains
of silver. The unexposed silver bromide is
removed by treating the film with sodium
thiosulfate. Soluble sodium silver thiosulfate
(Na3Ag(S2O3)2) is produced.
AgBr(s) 2Na2S2O3(aq) 0
Na3Ag(S2O3)2(aq) NaBr(aq)
44.01 g CO
__
2
1 mol CO2
191.0 g CO2 produced
70. Car Battery Car batteries use lead, lead(IV)
oxide, and a sulfuric acid solution to produce an
electric current. The products of the reaction are
lead(II) sulfate in solution and water.
a. Write the balanced chemical equation for
this reaction.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
b. If the mass of the ore from which the gold
Balance the equation, and determine the mass
of CO2 produced from the combustion of
100.0 g of ethanol.
Pb(s) PbO2(s) 2H2SO4(aq) 0
2PbSO4(aq) 2H2O(l)
b. Determine the mass of lead(II) sulfate
produced when 25.0 g of lead reacts with an
excess of lead(IV) oxide and sulfuric acid.
2 mol PbSO
1 mol Pb
25 g Pb __ __
Determine the mass of Na3Ag(S2O3)2 produced
if 0.275 g of AgBr is removed.
0.275 g AgBr 1 mol AgBr
__
187.77 g AgBr
1.46 103 mol AgBr
1.46 103 mol AgBr 1.46 103
1 mol Na Ag(S O )
__
3
2
3 2
1 mol AgBr
mol Na3Ag(S2O3)2
1.46 103 mol Na3Ag(S2O3)2
401.12 g Na3Ag(S2O3)2
0.587 g Na3Ag(S2O3)2
1 mol Na3Ag(S2O3)2
___
Section 11.3
4
207.2 g Pb
1 mol Pb
303.23 g PbSO4
73.2 g PbSO4
1 mol PbSO4
__
71. To extract gold from its ore, the ore is treated
with sodium cyanide solution in the presence of
oxygen and water.
4Au(s) 8NaCN(aq) O2(g) 2H2O(l) 0
4NaAu(CN)2(aq) 4NaOH(aq)
a. Determine the mass of gold that can be
extracted if 25.0 g of sodium cyanide is
used.
25.0 g NaCN 1 mol NaCN
4 mol Au
__
__
49.01 g NaCN
196.97 g Au
50.2 g Au
1 mol Au
__
Solutions Manual
8 mol NaCN
Mastering Concepts
73. How is a mole ratio used to find the limiting
reactant?
The actual mole ratio of reactants from the
chemical equation is compared to the mole ratio
determined from the given quantities.
74. Explain why the statement “the limiting
reactant is the reactant with the lowest mass” is
incorrect.
The limiting reactant is the reactant that produces
the lowest number of moles of product. Mass
does not determine the limiting reactant, but the
number of moles.
Chemistry: Matter and Change • Chapter 11
223
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CHAPTER
SOLUTIONS MANUAL
77. Nickel-Iron Battery In 1901, Thomas Edison
invented the nickel-iron battery. The following
reaction takes place in the battery.
M and circles to represent Element N.
a. Write the balanced equation for the reaction.
3M2 N2 0 2M3N
b. If each square represents 1 mol of M and
each circle represents 1 mol of N, how many
moles of M and N were present at the start
of the reaction?
6 moles of element M (in the form of 3 moles
of M2) and 6 moles of element N (likewise,
3 moles of N2)
c. How many moles of product form? How
many moles of M and N are unreacted?
2 moles of M3N form with 2 moles of N2
unreacted (4 total moles of element N)
How many mol of Fe(OH)2 are produced when
5.00 mol of Fe and 8.00 mol of NiO(OH) react?
According to the balanced equation, two moles
of NiO(OH) react with each mole of Fe. So, 4 mol
Fe react with 8.00 mol NiO(OH), leaving 1.00 mol
Fe in excess. For each mole of Fe that reacts, one
mole of Fe(OH)2(s) is produced. Because 4.0 mol
Fe reacts, 4.0 mol Fe(OH)2 is produced.
78. One of the few xenon compounds that form is
cesium xenon heptafluoride (CsXeF7). How
many moles of CsXeF7 can be produced from
the reaction of 12.5 mol of cesium fluoride with
10.0 mol of xenon hexafluoride?
CsF(s) XeF6(s) 0 CsXeF7(s).
10.0 mol XeF6 d. Identify the limiting reactant and excess
reactant.
7
1 mol XeF6
7
79. Iron Production Iron is obtained commer-
M2 is the limiting reactant and N2 is the excess
reactant.
Mastering Problems
76. The reaction between ethyne (C2H2) and
hydrogen (H2) is illustrated in Figure 11.13.
The product is ethane (C2H6). Which is the
limiting reactant? Which is the excess reactant?
Explain.
→
+
1 mol CsXeF
__
10.0 mol CsXeF
+
cially by the reaction of hematite (Fe2O3) with
carbon monoxide. How many grams of iron
are produced if 25.0 mol of hematite react with
30.0 mol of carbon monoxide?
Fe2O3(s) 3CO(g) 0 2Fe(s) 3CO2(g)
According to the balanced equation, 1 mole of
hematite reacts with 3 moles of carbon monoxide.
Since 25.0 mol of hematite would require
75.0 mol CO but only 30.0 mol are available, CO is
the limiting reactant.
30.0 mol CO 55.85 g Fe
2 mol Fe
_
_
3 mol CO
1 mol Fe
1120 g Fe
Ethyne
Hydrogen
Ethane
Ethyne
Hydrogen is limiting; ethyne is the excess
reactant. One mol of ethyne is left over.
224
Chemistry: Matter and Change • Chapter 11
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
75. Figure 11.12 uses squares to represent Element
Fe(s) 2NiO(OH)(s) 2H2O(l) 0
Fe(OH)2(s) 2Ni(OH)2(aq)
11
CHAPTER
SOLUTIONS MANUAL
80. The reaction of chlorine gas with solid
82. Lithium reacts spontaneously with bromine to
phosphorus (P4) produces solid phosphorus
pentachloride. When 16.0 g of chlorine reacts
with 23.0 g of P4, which reactant is limiting?
Which reactant is in excess?
produce lithium bromide. Write the balanced
chemical equation for the reaction. If 25.0 g of
lithium and 25.0 g of bromine are present at the
beginning of the reaction, determine
10Cl2(g) P4(s) 0 4PCl5(s)
2Li(s) Br2(l) 0 2LiBr(s)
1 mol CI
__
0.226 mol CI
23.0 g P4 a. the limiting reactant.
2
16.0 g Cl2 2
70.90 g CI2
25.0 g Li 1 mol P
__
0.185 moles
1 mol Li
_
3.60 mol Li
6.94 g Li
4
123.88 g P4
According to the balanced equation, Cl2 reacts
with P4 in a ten-to-one ratio.
0.226 mol Cl2 1 mol P
_
4
10 mol CI2
0.0226 mol P4 needed
25.0 g Br2 1 mol Br
__
0.156 mol Br
2
159.80 g Br2
Actual ratio of mol Li to mol Br2 is 3.60 mol
Li/0.156 Br2 or 23:1. Only 2 mol Li are required
for 1 mol Br2. Thus, Br2 is the limiting reactant.
b. the mass of lithium bromide produced.
Cl2 is limiting reactant; P4 is in excess.
0.156 mol Br2 2 mol LiBr
_
0.312 mol LiBr
0.312 mol LiBr 86.84 g LiBr
__
27.1 g LiBr
81. Alkaline Battery An alkaline battery
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
produces electrical energy according to this
equation.
Zn(s) 2MnO2(s) H2O(l) 0
Zn(OH)2(s) Mn2O3(s)
a. Determine the limiting reactant if 25.0 g of
Zn and 30.0 g of MnO2 are used.
25.0 g Zn 1 mol Zn
_
0.380 mol Zn
65.3 g Zn
30.0 g MnO2 1 mol MnO
__
2
According to the balanced equation, MnO2
reacts with Zn in a two-to-one ratio. In the
reaction, the ratio is 1:1.1 or 0.345/0.380.
MnO2 is the limiting reactant.
b. Determine the mass of Zn(OH)2 produced.
1 mol Zn(OH)
__
2
2 mol MnO2
99.39 g Zn(OH)
__
17.1 g Zn(OH)
2
1 mol Zn(OH)2
2
1 mol Br2
1 mol LiBr
c. the excess reactant and the excess mass.
Li is in excess.
0.156 mol Br2 2 mol Li
_
0.312 mol Li used
1 mol Br2
3.60 mol Li 0.312 mol Li used 3.29 mol Li
remaining
3.29 mol Li 86.92 g MnO2
0.345 mol MnO2
0.345 mol MnO2 2
6.94 g Li
_
22.8 g Li remaining
1 mol Li
Section 11.4
Mastering Concepts
83. What is the difference between actual yield and
theoretical yield?
Actual yield is the amount of product actually
obtained experimentally. Theoretical yield is the
amount of product predicted by a stoichiometric
calculation.
84. How are actual yield and theoretical yield
determined?
Actual yield is determined through
experimentation. Theoretical yield is calculated
from a given reactant or the limiting reactant.
Solutions Manual
Chemistry: Matter and Change • Chapter 11
225
11
SOLUTIONS MANUAL
85. Can the percent yield of a chemical reaction be
more than 100%? Explain your answer.
No, you cannot produce more product than the
theoretical yield, which is determined from the
starting reactants.
86. What relationship is used to determine the
percent yield of a chemical reaction?
Mastering Problems
90. Ethanol (C2H5OH) is produced from the
fermentation of sucrose (C12H22O11) in the
presence of enzymes.
C12H22O11(aq) H2O(g) 0
4C2H5OH(l) 4CO2(g)
Determine the theoretical yield and the percent
yield of ethanol if 684 g of sucrose undergoes
fermentation and 349 g of ethanol is obtained.
actual yeild
__
100 percent yield
theoretical yield
87. What experimental information do you need in
order to calculate both the theoretical and the
percent yield of any chemical reaction?
The quantity of one reactant and the actual yield
of the product
88. A metal oxide reacts with water to produce a
metal hydroxide. What additional information
would you need to determine the percent yield
of metal hydroxide from this reaction?
the mass of one substance in the reaction and the
actual mass of metal hydroxide produced
89. Examine the reaction represented in Figure 11.14.
Determine if the reaction went to completion.
Explain your answer, and calculate the percent
yield of the reaction.
Element A
Element B
theoretical yield:
684 g C12H22O11 1 mol C H O
__
12 22
11
342.23 g C12H22O11
46.07 g C H OH
4 mol C H OH
__
__
2 5
2 5
1 mol C12H22O11
369 g C2H5OH
% yield 1 mol C2H5OH
actual yield
__
100
theoretical yield
349 g
100 94.6%
369 g
_
91. Lead(II) oxide is obtained by roasting galena,
lead(II) sulfide, in air. The unbalanced equation is:
PbS(s) O2(g) 0 PbO(s) SO2(g)
a. Balance the equation, and determine the
theoretical yield of PbO if 200.0 g of PbS is
heated.
2PbS 3O2 0 2PbO 2SO2
Theoretical yield 200.0 g PbS The reaction did not go to completion. Using
squares to represent Element A and circles to
represent Element B, the initial products would
have yielded four AB2 particles, but only three
were produced. There are enough remaining
unreacted A and B particles to produce one more
AB2 particle. The percent yield is 75%.
226
Chemistry: Matter and Change • Chapter 11
1 mol PbS
__
239.27 g PbS
223.19 g PbO
2 mol PbO
__ __ 186.6 g PbO
2 mol PbS
1 mol PbO
b. What is the percent yield if 170.0 g of PbO
is obtained?
% yield 170.0 g
_
100 91.10%
186.6 g
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CHAPTER
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11
SOLUTIONS MANUAL
92. Upon heating, calcium carbonate (CaCO3)
c. What is the theoretical yield of H2SiF6?
decomposes to produce calcium oxide (CaO)
and carbon dioxide (CO2).
a. Determine the theoretical yield of CO2 if
235.0 g of CaCO3 is heated.
2.00 mol HF 0.333 mol H2SiF6 1 mol CaCO
__
100.06 g CaCO
43.99 g CO
1
mol
CO
__ __ 103 g CO
1 mol CO2
2
b. What is the percent yield of CO2 if 97.5 g of
actual yield
__
100
theoretical yield
97.5 g CO
__ 100 94.7%
% yield 45.8 g H SiF
__
100 95.4% yield
2
6
48.0 H2SiF6
94. Van Arkel Process Pure zirconium is
% yield 2
103 g CO2
93. Hydrofluoric acid solutions cannot be stored in
glass containers because HF reacts readily with
silica dioxide in glass to produce hexafluorosilicic acid (H2SiF6).
SiO2(s) 6HF(aq) 0 H2SiF6(aq) 2H2O(l)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
6
d. What is the percent yield?
CO2 is collected?
40.0 g SiO2 and 40.0 g HF react to yield 45.8 g
H2SiF6.
a. What is the limiting reactant?
40.0 g SiO2 2
1 mol H2SiF6
48.0 g H2SiF6 is the theoretical yield.
3
2
2
6
144.11 g H SiF
__
48.0 g H2SiF6
3
1 mol CaCO3
2
6 mol HF
0.333 mol H2SiF6
CaCO3(s) 0 CaO(s) CO2(g)
235 g CaCO3 1 mol H SiF
__
__ 0.666 mol SiO
1 mol SiO2
2
60.09 g SiO2
obtained using the two-step Van Arkel process.
In the first step, impure zirconium and iodine
are heated to produce zirconium iodide (ZrI4).
In the second step, ZrI4 is decomposed to
produce pure zirconium.
ZrI4(s) 0 Zr(s) 2I2(g)
Determine the percent yield of zirconium if
45.0 g of ZrI4 is decomposed and 5.00 g of pure
Zr is obtained.
theoretical yield 45.0 g ZrI4 1 mol ZrI
__
4
598.84 g ZrI4
91.22 g Zr
1 mol Zr
_
_ 6.85 g of Zr
1 mol ZrI4
1 mol Zr
actual yield
__
100
theoretical yield
5.00 g Zr
_ 100 73.0%
% yield 1 mol HF
40.0 g HF __ 2.00 mol HF
20.01 g HF
Ratio of HF to SiO2 in the balanced equation is
6:1. Actual ratio of HF to SiO2 is
2.00 mol HF
or 3.00:1. Thus, HF is the
0.666 mol SiO2
limiting reactant.
6.85 g Zr
__
b. What is the mass of the excess reactant?
2.00 mol HF 1 mol SiO
__
0.333 mol SiO
2
2
6 mol HF
0.666 mol SiO2 available 0.333 mol SiO2 used
0.333 mol SiO2 in excess
0.333 mol SiO2 60.09 g SiO
__
2
1 mol SiO2
20.0 g SiO2 in excess
Solutions Manual
Chemistry: Matter and Change • Chapter 11
227
11
SOLUTIONS MANUAL
According to the balanced equation, Ca3(PO4)2
reacts with SiO2 in a one-to-three ratio. In this
reaction, SiO2 is in excess and 0.8060 mol of
Ca3(PO4)2 react.
95. Methanol, wood alcohol, is produced when
carbon monoxide reacts with hydrogen gas.
CO 2H2 0 CH3OH
When 8.50 g of carbon monoxide reacts with
an excess of hydrogen, 8.52 g of methanol is
collected. Complete Table 11.4, and calculate
the percent yield for this reaction.
CO(g)
CH3OH(l)
Mass
8.50 g
9.71 g
Molar mass
28.01 g/mol
32.05 g/mol
Moles
0.303 mol
0.303 mol
0.303 mol CO (1 mol CH3OH/1 mol CO)
0.303 mol CH3OH
Percent yield (8.52 g/9.71 g) 100 87.7% yield
96. Phosphorus (P4) is commercially prepared
by heating a mixture of calcium phosphate
(CaSiO3), sand (SiO2), and coke (C) in an
electric furnace. The process involves two
reactions.
10
49.92 g of P4
1 mol P
123.88 g P
__
__
4
4
1 mol P4O10
1 mol P4
Step 4 is to determine the percent yield, given
actual yield is 49.92 g.
actual yield
__
100
theoretical yield
45.0 g P
_ 100 90.1%
4
49.92 g P4
97. Chlorine forms from the reaction of hydro-
chloric acid with manganese(IV) oxide. The
balanced equation is:
Calculate the theoretical yield and the percent
yield of chlorine if 86.0 g of MnO2 and 50.0 g
of HCl react. The actual yield of Cl2 is 20.0 g.
Step 1: The balanced equation is given.
P4O10(g) 10C(s) 0 P4(g) 10CO(g)
The P4O10 produced in the first reaction reacts
with an excess of coke (C) in the second reaction. Determine the theoretical yield of P4 if
250.0 g of Ca3(PO4)2 and 400.0 g of SiO2 are
heated. If the actual yield of P4 is 45.0 g, determine the percent yield of P4.
Step 1 is to determine the excess quantity in
equation #1.
228
4
2 mol Ca3(PO4)2
MnO2 4HCl 0 MnCl2 Cl2 2H2O
2Ca3(PO4)2(s) 6SiO2(s) 0
6CaSiO3(l) P4O10(g)
0.8060 mol
1 mol P O
__
% yield 0.304 mol CH3OH (32.05 g CH3OH/1 mol CH3OH)
9.71 g CH3OH
1 mol SiO
__
6.657 mol of SiO
2
2
60.08 g SiO2
1 mol Ca (PO )
__
3
0.4030 mol P4O10
0.4030 mol P4O10 8.50 g CO (1 mol CO/28.01 g CO) 0.303 mol CO
250.0 g Ca3(PO4)2 0.8060 mol Ca3(PO4)2 Step 3 is to determine amount of P4 produced in
step 2 from 0.4030 mol P4O10.
Methanol Reaction Data
400.0 g SiO2 Step 2 is to determine amount of P4O10 produced.
MnO2 (s) 4HCl (aq) 0 MnCl2 (aq) Cl2 (g) 2H2O (l)
Step 2 is to determine which reactant is in excess.
86.0 g MnO2 50.0 g HCl 1 mol MnO
__
0.989 mol MnO
2
2
86.94 g MnO2
1 mol HCI
__
1.37 mol HCI
36.46 g HCI
According to the balanced equation, MnO2
reacts with HCl in a one-to-four ratio. In this
reaction, the ratio is 0.989 mol/1.37 mol or 1:1.38.
Therefore MnO2 is in excess, and HCl is the
limiting reactant.
4 2
310.17 g Ca3(PO4)2
Chemistry: Matter and Change • Chapter 11
Solutions Manual
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CHAPTER
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11
SOLUTIONS MANUAL
Step 3 is to determine the grams of Cl2 produced.
1.37 mol HCl a. Write the balanced chemical equation for
the formation of nitrogen dioxide from
nitrogen oxide.
1 mol CI
70.90 g CI
_
__
2
4 mol HCI
24.3 g CI2.
2
1 mol CI2
2NO(g) O2(g) 0 2NO2(g)
Step 4 is to determine percent yield, given that
20.0 g of Cl2 is actually formed.
b. What mole ratio would you use to convert
from moles of nitrogen oxide to moles of
nitrogen dioxide?
_ 100 82.3%
20.0 g CI2
24.3 g CI2
2 mol NO
__
2
2 mol NO
Mixed Review
98. Ammonium sulfide reacts with copper(II)
nitrate in a double replacement reaction. What
mole ratio would you use to determine the
moles of NH4NO3 produced if the moles of
CuS are known?
102. Electrolysis Determine the theoretical
and percent yield of hydrogen gas if 36.0 g
of water undergoes electrolysis to produce
hydrogen and oxygen and 3.80 g of hydrogen
is collected.
(NH4)2S Cu(NO3)2 0 2NH4NO3 CuS
2H2O(l) 0 H2(g) O2(g)
2 mol NH NO
__
36.0 g H2O (1 mol H2O/18.02 g H2O) 2.00 mol
4
3
1 mol CuS
theoretical yield 2.00 mol H2O (CaNCN) can be used as a nitrogen source
for crops. To obtain this compound, calcium
carbide is reacted with nitrogen at high
temperatures.
CaC2(s) N2(g) 0 CaNCN(s) C(s)
What mass of CaNCN can be produced if
7.50 mol of CaC2 reacts with 5.00 mol of N2?
5.00 mol N2 80.11 g CaNCN
1 mol CaNCN
__
__
1 mol N2
1 mol CaNCN
401 g CaNCN
100. When copper(II) oxide is heated in the presence
of hydrogen gas, elemental copper and water
are produced. What mass of copper can be
obtained if 32.0 g of copper(II) oxide is used?
CuO H2 0 Cu H2O
32.0 g CuO 1 mol Cu
1 mol CuO
__
__ 79.55 g CuO
1 mol CuO
63.55 g Cu
__
25.6 g Cu
1 mol Cu
101. Nitrogen oxide is present in urban pollution,
but it immediately converts to nitrogen dioxide
as it reacts with oxygen.
Solutions Manual
2.02 g H
_
4.04 g H
2
2 mol H2O
2
2
1 mol H2
% yield 3.80 g H
_
100 94.1%
2
4.04 g H2
103. Iron reacts with oxygen as shown.
Mass of Fe2O3 Formed From Burning Fe
Mass of Fe2O3 (g)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
99. Fertilizer The compound calcium cyanamide
1 mol H
__
30
20
10
0
0
5
10
15
20
25
30
35
Mass of Fe (g)
4Fe(s) 1 3O2(g) 0 2Fe2O3(s)
Different amounts of iron were burned in a
fixed amount of oxygen. For each mass of iron
burned, the mass of iron(II) oxide formed was
plotted on the graph shown in Figure 11.15.
Why does the graph level off after 25.0 g of
iron is burned? How many moles of oxygen
are present in the fixed amount?
Chemistry: Matter and Change • Chapter 11
229
11
SOLUTIONS MANUAL
sulfate. Determine the actual yield of copper(II)
sulfate. Divide the actual yield by the theoretical
yield and multiply by 100 to determine percent
yield of copper(II) sulfate.
The graph levels off because the oxygen limits
the reaction at that point.
35.0 g Fe2O3 0.329 mol O2
1 mol Fe O
3 mole O
__
__
2
3
159.7 g Fe2O3
2
2 mole Fe2O3
107. Apply Concepts When a campfire begins
to die down and smolder, the flame can be
rekindled if you start to fan it. Explain in terms
of stoichiometry why the fire begins to flare up
again.
Think Critically
104. Analyze and Conclude In an experiment,
you obtain a percent yield of product of 108%.
Is such a percent yield possible? Explain.
Assuming that your calculation is correct, what
reasons might explain such a result?
When you fan the flame, additional oxygen is
added and the remaining coals can burn.
108. Apply Concepts Students conducted a
lab to investigate limiting and excess reactants. The students added different volumes
of sodium phosphate solution (Na3PO4) to a
beaker. They then added a constant volume of
cobalt(II) nitrate solution (Co(NO3)2), stirred
the contents, and allowed the beakers to sit
overnight. The next day, each beaker had a
purple precipitate formed at the bottom. The
students decanted the supernatant from each
beaker, divided it into two samples, and added
one drop of sodium phosphate solution to one
sample and one drop of cobalt(II) nitrate solution to the second sample. Their results are
shown in Table 11.5.
No, percent yields cannot be greater than 100%.
High results could mean the product was not
completely dry, or it was contaminated.
105. Observe and Infer Determine whether
each reaction depends on a limiting reactant.
Explain why or why not, and identify the
limiting reactant.
a. Potassium chlorate decomposes to form
potassium chloride and oxygen.
No, because there is only one reactant.
b. Silver nitrate and hydrochloric acid react to
produce silver chloride and nitric acid.
Yes, because there are two reactants. Not
enough information is given to identify
which is the limiting reactant.
Reaction
with Drop
of Na3PO4
Reaction
with
Drop of
Co(NO3)2
Trial
Volume
Na3PO4
Volume
Co(NO3)2
ment that can be used to determine the percent
yield of anhydrous copper(II) sulfate when
copper(II) sulfate pentahydrate is heated to
remove water.
1
5.0 mL
10.0 mL
purple
no
precipitate reaction
2
10.0 mL
10.0 mL
no
reaction
purple
precipitate
Obtain and record the mass of an empty
evaporating dish. Add 2.00 g of copper(II)
sulfate pentahydrate and obtain and record the
mass of the hydrate and evaporating dish. Heat
the dish gently for 5 minutes, then strongly for
5 minutes to drive off the water. Cool the dish
and remass. Record. Determine the mass of the
anhydrous copper sulfate. Using the equation
CuSO4·5H2O 0 CuSO4 5H2O and the initial
mass of the copper(II) sulfate pentahydrate,
determine the theoretical yield of copper(II)
3
15.0 mL
10.0 mL
no
reaction
purple
precipitate
4
20.0 mL
10.0 mL
no
reaction
purple
precipitate
106. Design an Experiment Design an experi-
230
Reaction Data for Co(NO3)2 and Na3PO4
Chemistry: Matter and Change • Chapter 11
a. Write a balanced chemical equation for the
reaction.
2Na3PO4 (aq) 3Co(NO3)2 (aq) 0 Co3(PO4)2
(s) 6NaNO3 (aq)
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CHAPTER
CHAPTER
11
b. Based on the results, identify the limiting
b. Write balanced equations for the steps of
reactant and the excess reactant for each
trial.
the reaction.
Trial 1: Na3PO4 is limiting, Co(NO3)2 is in
excess because the addition of Na3PO4
caused an additional reaction.
VO2 2H2 0 V 2H2O
Trials 2-4: Co(NO3)2 is limiting, Na3PO4 is in
excess because the addition of Co(NO3)2
caused and additional reaction.
Challenge Problem
109. When 9.59 g of a certain vanadium oxide is
heated in the presence of hydrogen, water and
a new oxide of vanadium are formed. This
new vanadium oxide has a mass of 8.76 g.
When the second vanadium oxide undergoes
additional heating in the presence of hydrogen,
5.38 g of vanadium metal forms.
a. Determine the empirical formulas for the
two vanadium oxides.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
SOLUTIONS MANUAL
V:
5.38 g
__
0.106 mol
O:
4.21 g
__
0.263 mol
50.94 g/mol
15.999 g/mol
Divided the smallest molar quantity.
0.106 mol
_
1
0.106 mol
0.263 mol
_
2.48 艐 2.5
0.106 mol
Multiply by the mole ratio by 2.
2 (1 mol V: 2.5 mol O) V2O5
V:
O:
5.38 g
__
0.106 mol
50.94 g/mol
3.38 g
__
0.211 mol
15.999 g/mol
Divided the smallest molar quantity.
0.106 mol
_
1
0.106 mol
0.211 mol
_
1.99 艐 2
V2O5 H2 0 2VO2 H2O
c. Determine the mass of hydrogen needed to
complete the steps of this reaction.
9.59 g V2O5
1 mol H2
__
__
181.88 g/mol
1 mol V2O5
2.016 g H2
__
0.106 g H2
1 mol H2
2 mol H2
8.76 g VO2
__
_
82.94 g/mol
1 mol VO2
2.016 g H2
0.426 g H2
1 mol H2
__
Total hydrogen mass 0.106 g 0.426 g 0.532 g H2
Cumulative Review
110. You observe that sugar dissolves more quickly
in hot tea than in iced tea.You state that higher
temperatures increase the rate at which sugar
dissolves in water. Is this statement a hypothesis or a theory? Why? (Chapter 1)
a hypothesis, because it is based only on
observation, not on data
111. Write the electron configuration for each of
the following atoms. (Chapter 5)
a. fluorine
[He]2s22p5
b. aluminum
[Ne]3s23p1
c. titanium
[Ar]4s23d2
d. radon
[Xe]6s24f145d106p6
0.106 mol
1 mol V: 2 mol O VO2
Solutions Manual
Chemistry: Matter and Change • Chapter 11
231
11
CHAPTER
SOLUTIONS MANUAL
112. Explain why the gaseous nonmetals exist
b. What is the empirical formula of butanoic
as diatomic molecules, but other gaseous
elements exist as single atoms. (Chapter 8)
acid?
Diatomic molecules achieve a noble gas electron
configuration by forming covalent bonds. The
monoatomic gases already have noble gas
electron configurations.
9.1 g H (1 mol/1.01 g) 9.0 mol
113. Write a balanced equation for the reaction of
potassium with oxygen. (Chapter 9)
54.5 g C (1 mol/12.01 g) 4.54 mol
36.4 g O (1 mol/16.00 g ) 2.28 mol
4.54 mol C/2.28 mol O 2
9.0 mol H/2.28 mol O 4
2.28 mol O/2.28 mol O 1
4K(s) O2(g) 0 2K2O(s)
The empirical formula is C2H4O.
114. What is the molecular mass of UF6? What is
the molar mass of UF6? (Chapter 10)
Additional Assessment
352.02 amu, 352.02 g/mol
116. Air Pollution Research the air pollutants
115. Figure 11.16 gives percent composition data
for several organic compounds. (Chapter 10)
Percent Composition of
Some Organic Compounds
%C
%H
%O 40.0
Percent by mass
52.2
50
40
53.3
54.5
54.5
36.4
34.8
36.4
30
13.0
10
6.7
Ethanol
9.1
9.1
Formaldehyde Acetaldehyde Butanoic acid
Compound name
a. How are the molecular and empirical
formulas of acetaldehyde and butanoic acid
related?
They are whole number multiples of one
another.
232
Answers will vary. Common pollutants are SO2,
NO, NO2 and O3. Check that stoichiometric
calculations account for a decrease in the
pollutant.
117. Haber Process The percent yield of
20
0
produced by combustion of gasoline in internal
combustion engines. Discuss the common
pollutants and the reaction that produces them.
Show, through the use of stoichiometry, how
each pollutant could be reduced if more people
used mass transit.
Chemistry: Matter and Change • Chapter 11
ammonia produced when hydrogen and
nitrogen are combined under ordinary
conditions is extremely small. However, the
Haber Process combines the two gases under a
set of conditions designed to maximize yield.
Research the conditions used in the Haber
Process, and find out why the development of
the process was of great importance.
Answers will vary. Make sure the equation, N2(g)
3H2(g) 2NH3(g) 92 kJ, is included. The aim
of the Haber process was to control a reaction
so that a large amount of a useful product
was yielded quickly. The process was of great
importance because the Germans had to come
up with a nitrogen compound that could be
produced in large amounts.
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238.03 6(18.998) 352.02
CHAPTER
11
SOLUTIONS MANUAL
2 mol C6H4(OH)2
__
from eqauation
Document-Based Questions
4 mol H2O2
Bombardier Beetles Many insects secrete hydrogen
peroxide (H2O2) and hydroquinone C6H4(OH)2.
Bombardier beetles take this a step further by
mixing these chemicals with a catalyst. The result
is an exothermic chemical reaction and a spray of
hot, irritating chemicals for any would-be predator.
Researchers hope to use a similar method to reignite
aircraft turbine engines.
H2O2 is the limiting reactant.
119. What is the excess reactant, and how many
milligrams are in excess?
2 mol C H (OH)
__
4 mol H O
110.12 g C H (OH)
1 mg
__
__
2
6 4
100.0 mg 80.9 mg 19.1 mg of C6H4(OH)2 in
excess
120. How many milligrams of benzoquinone will
be produced?
1.47 103 mol H2O2 H2O2
H2O O2 Energy
108.09 g C6H4O2
__
O
C6H4(OH)2
Hydroquinone
C6H4O2
Benzoquinone
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
118. If the bombardier beetle stores 100.0 mg of
hydroquinone (C6H4(OH)2) along with
50.0 mg of hydrogen peroxide (H2O2), what is
the limiting reactant?
Balanced reaction:
2C6H4(OH)2 4H2O2 0 2C6H4O2 6H2O O2
100 mg
50 mg
2 mol C6H4O2
__
4 mol H2O2
1
mg
__
1 mol C6H4O2
1 103 g
79.4 mg C6H4O2 produced
Catalyst
OH
2
2
O
OH
2
1 mol C6H4(OH)2
1 103 g
80.9 mg C6H4(OH)2 used
Figure 11.17 below shows the chemical reaction that
results in the bombardier beetle’s defensive spray.
Data obtained from: Becker, Bob. April 2006. ChemMatters.
24: no.2
6 4
1.47 103 mol H2O2 Standardized Test Practice
pages 398–399
1. Stoichiometry is based on the law of
a.
b.
c.
d.
constant mole ratios.
Avogadro’s constant.
conservation of energy.
conservation of mass.
d
? mg
C6H4(OH)2 H2O2 0 C6H4O2 H2O O2
energy
Use the graph below to answer Questions 2–5.
Supply of Various Chemicals
in Dr. Raitano’s Laboratory
__
3
100.0 mg C6H4(OH)2 1 10 g
1 mg
1
mol
__
9.08 104 mol
110.00 g C6H4(OH)2
C6H4(OH)2
NaCl
700.0 g
Na2CO3
500.0 g
Ca(OH)2
300.0 g
__ __
3
1 mol
50.0 mg H2O2 1 10 g 1 mg
34.02 g H2O2
1.47 103 mol H2O2
___ 9.08 104 mol C6H4(OH)2
AgNO3
100.0 g
KClO3
200.0 g
NaH2PO4
350.0 g
1.47 103 mol H2O2
0.618 mol C6H4(OH)2
__
from reaction vs.
1 mol H2O2
Solutions Manual
Chemistry: Matter and Change • Chapter 11
233
11
SOLUTIONS MANUAL
2. Pure silver metal can be made using the reac-
Cu(s) 2AgNO3(aq) 0
2Ag(s) Cu(NO3)2(aq)
2KClO3(s) 0 2KCl(s) 3O2(g)
How many grams of copper metal will be
needed to use up all of the AgNO3 in
Dr. Raitano’s laboratory?
a. 18.70 g
b. 37.3 g
c. 74.7 g
d. 100 g
a
100.0 g AgNO3 __
1 mol AgNO
1mol Cu
__
__
3
169.88 g AgNO3
63.55w g Cu
18.70 g Cu
1 mol Cu
2 mol AgNO3
of manufacturing sodium hydroxide. The equation for this process is as follows.
Na2CO3(aq) Ca(OH)2(aq) 0
2NaOH(aq) CaCO3(s)
Using the amounts of chemicals available in
Dr. Raitano’s lab, what is the maximum number
of moles of NaOH that can be produced?
a. 4.05 mol
b. 4.72 mol
c. 8.097 mol
d. 9.43 mol
c
Find the limiting reactant.
1 mol Na CO
__
2
__ 1 mol Ca2 (OH)2
74.10 g Ca(OH)2
4.049 mol Ca(OH)2
Ca(OH)2 is the limiting reactant.
4.049 mol Ca(OH)2 b
_1 (200.0 g KCIO3) 100.0 g KCIO3
2
theoretical yield 100.0 g KCIO3 1 mol KCIO
__
3
122.5 g KCIO3
32.00 g O
3 mol O
__
__ 39.17 g O
2
2
2 mol KCIO3
percent 2
1 mol O2
actual yield
12.8 g
__
100 _
39.17 g
theoretical yield
100 32.7%
5. Sodium dihydrogen pyrophosphate
(Na2H2P2O7), more commonly known as
baking powder, is manufactured by heating
NaH2PO4 at high temperatures:
2NaH2PO4(s) 0 Na2H2P2O7(s) H2O(g)
If 444.0 g of Na2H2P2O7 is needed, how much
more NaH2PO4 will Dr. Raitano have to buy to
make enough Na2H2P2O7?
a. 0.00 g
b. 94.0 g
c. 130.0 g
d. 480 g
3
106.00 g Na2CO3
4.717 mol Na2CO3
300.0 g Ca(OH)2
If half of the KClO3 in the lab is used and
12.8 g of oxygen gas is produced, what is the
percent yield of this reaction?
a. 12.8%
b. 32.7%
c. 65.6%
d. 98.0%
3. The LeBlanc process is the traditional method
500.0 g Na2CO3 4. Pure O2 gas can be generated from the decom-
position of potassium chlorate (KClO3):
tion shown below.
2 mol NaOH
__
c
1 mol Na H P O
__
221.94 g Na H P O
119.98 g NaH PO
2 mol NaH PO
__
__
2 2 2
444.0 g Na2H2P2O7 2
4
1 mol Na2H2P2O7
48.0 g NaH2PO4
7
2 2 2
2
7
4
1 mol Na2H2PO4
480.0 g 350.0 g 130.0 g
1 mol Ca(OH)2
8.097 mol NaOH
234
Chemistry: Matter and Change • Chapter 11
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
CHAPTER
CHAPTER
11
SOLUTIONS MANUAL
6. Red mercury(II) oxide decomposes at high
temperatures to form mercury metal and oxygen
gas:
2HgO(s) 0 2Hg(l) O2(g)
If 3.55 mol of HgO decomposes to form
1.54 mol of O2 and 618 g of Hg, what is the
percent yield of this reaction?
a. 13.2%
b. 42.5%
c. 56.6%
d. 86.8%
d
__
1 mol O2
theoretical yield O2 3.55 mol HgO 2 mol HgO
1.775 mol O2
percent yield 1.54 mol
_
100 86.8%
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
(CH3)2N2H2(l) 2N2O4(l) 0
3N2(g) 4H2O(g) 2CO2(g)
Because this reaction produces an enormous
amount of energy from a small amount of
reactants, it was used to drive the rockets on
the Lunar Excursion Modules (LEMs) of the
Apollo space program. If 18.0 mol of dinitrogen
tetroxide is consumed in this reaction, how
many moles of nitrogen gas will be released?
Set up a mole ratio:
27 moles N2
18
1
13 14 15 16 17
2
neously upon contact with dinitrogen tetroxide
(N2O4):
18 moles N2O4 1.775 mol
PERIODIC TABLE
Y
Y
Y
Y
Y
Y
9. Dimethyl hydrazine (CH3)2N2H2 ignites sponta-
W W
Y
Y 3 4 5 6 7 8 9 10 11 12 W W
Y Z Z Z Z Z Z Z Z Z Z W W
Y Z Z Z Z Z Z Z Z Z Z W W
Y Z Z Z Z Z Z Z Z Z Z W W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
W
Y
W
W
W
W
W
Y Y Z Z Z
X X X X X X X X X X X X X X
X X X X X X X X X X X X X X
7. Which elements tend to have the largest atomic
radius in their periods?
a. W
b. X
c. Y
d. Z
3 mole N
__
2
2 moles N2O4
3 moles N
__
2
2 moles N2O4
Use the table below to answer questions 10 and 11.
First Ionization Energy of Period 3 Elements
Element
Atomic Number
1st Ionization
Energy, kJ/mol
Sodium
11
496
Magnesium
12
736
Aluminum
13
578
Silicon
14
787
Phosphorus
15
1012
Selenium
16
1000
Chlorine
17
1251
Argon
18
1521
c
8. Elements labeled W have their valence electrons
in which sublevel?
a. s
b. p
c. d
d. f
b
Solutions Manual
Chemistry: Matter and Change • Chapter 11
235
11
CHAPTER
SOLUTIONS MANUAL
10. Plot the data from this data table. Place atomic
numbers on the x-axis.
Use the pictures below to answer Questions 13–17.
a.
Data should form an approximately linear
relationship with a few jagged edges, similar to
Figure 6.16 on page 191.
First Ionization
Energy (Kj/mol)
Atomic Number vs.
First Ionization Energy
b.
1600
1400
1200
1000
800
600
400
200
c.
10
12
14
16
18
Atomic Number
energy. How does ionization energy relate to
the number of valence electrons in an element?
d.
Ionization generally increases as you move across
a period or row in the periodic table. Elements
in the first few families have only 1 or 2 valence
electrons, which are relatively easy to remove
since this will result in a complete outer shell.
Elements on the right side of the periodic table
have very high ionization energies because their
outer shells are nearly filled, therefore making
it more likely for these elements to gain a few
electrons rather than lose many.
e.
12. How much cobalt(III) titanate (Co2TiO4), in
moles, is in 7.13 g of the compound?
a. 2.39 101 mol
b. 3.10 102 mol
c. 3.22 101 mol
d. 4.17 102 mol
13. Hydrogen sulfide displays this molecular shape.
a
14. Molecules with this shape have four shared
pairs of electrons and no lone pairs of electrons.
c
15. This shape of molecule is referred to as trigonal
planar.
b
b
7.13 g Co2TiO4 3.10 102 mol
1 mol Co TiO
__
0.0310 mol 2
4
229.74 g Co2TiO4
16. Carbon dioxide displays this molecular shape.
d
17. Molecules with this shape undergo sp2
hybridization.
b
236
Chemistry: Matter and Change • Chapter 11
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
11. Summarize the general trend in ionization