Chapter 7 Review, pages 346–351
Knowledge
1. (d)
2. (a)
3. (c)
4. (a)
5. (b)
6. (a)
7. (b)
8. (d)
9. (d)
10. (c)
11. False. The value that would complete the ratio below is 18.
3
7
=
?
42
12. False. The coefficients for the balanced chemical equation below are 3, 1, and 2.
3 H2(g) + N2(g) → 2 NH3(g)
13. True
14. False. Stoichiometric amounts in moles are multiples of the coefficients in a balanced
chemical equation.
15. True
16. False. When the supply of a reactant in a chemical reaction runs out and the reaction stops,
that reactant is said to be limiting.
17. True
18. False. If a butane, C4H10, tank were attached to a stove designed for propane, C3H8, the
air/fuel mixture would likely be somewhat richer than the ideal mixture.
19. False. The quantity that directly links the amount of limiting reagent in a reaction to the
amount of product is the mole ratio.
20. (a) (ii)
(b) (iv)
(c) (iii)
(d) (i)
21. (a) The ratio of the amount of HCl to the amount of Al is 3:1.
(b) The ratio of the amount of Al to the amount of AlCl3 is 1:1.
(c) The ratio of the amount of H2 to the amount of Al is 3:2.
22. (a) Given: 3 CuO(s) + 2 Al(s) → 3 Cu(l) + Al2O3(s) + heat
amount of aluminum, nAl = 0.4 mol
Required: amount of copper, nCu
Solution:
3 CuO(s) + 2 Al(s) → 3 Cu(l) + Al2O3(s) + heat
[÷ 5]
0.6 CuO(s) + 0.4 Al(s) → 0.6 Cu(l) + 0.2 Al2O3(s) + heat
Statement: When 0.4 mol of aluminum is used, then 0.6 mol of copper is produced.
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Chapter 7: Stoichiometry in Chemical Reactions
7-2
(b) Given: 3 CuO(s) + 2 Al(s) → 3 Cu(l) + Al2O3(s) + heat
amount of Cu, nCu = 0.2 mol
Required: amount of copper oxide, nCuO
Solution:
3 CuO(s) + 2 Al(s) → 3 Cu(l) + Al2O3(s) + heat
[÷ 15]
0.2 CuO(s) + 0.13 Al(s) → 0.15 Cu(l) + 0.067 Al2O3(s) + heat
Statement: When 0.2 mol of copper is produced, 0.2 mol of copper oxide is required.
23. Nitrogen gas is found in the atmosphere and also fills up sodium azide type airbags.
24. An example of the chemical reaction is an antacid tablet reducing acidity in the stomach.
25. When the combustion of methane gas is incomplete, a yellow-orange flame results.
26. Competing reactions consume some of the reactants to form an alternative product, thus
preventing the full consumption of the reactants to form the desired product. As a result,
competing reactions lower the percentage yield.
Understanding
27. The coefficients in a balanced chemical equation describe the mole ratio relating the amounts
of reactants and products, thus providing relative amounts for many different specific cases.
Similarly, a recipe can provide relative amounts of ingredients needed for making a dish.
However, the coefficients in a chemical equation only provide ratios of amounts, whereas recipes
may mix masses/weights or volumes with numbers of entities, such as stating cups of flour to go
with a number of eggs.
28. (a) In balanced chemical equations, the number of molecules is sometimes conserved. For
example, in the balanced equation H2 + Cl2 → 2 HCl, the total number of reactant molecules
equals the total number of product molecules. However, in the balanced equation
3 H2 + N2 → 2 NH3, the total number of reactant molecules is greater than the total number of
product molecules.
(b) In balanced chemical equations, the number of atoms is always conserved because the
equations are balanced to ensure that this is true.
29. The researcher can bubble a gas suspected to contain carbon dioxide through limewater. If
the gas contains carbon dioxide, a precipitate of calcium carbonate will form, turning the
limewater cloudy.
30. Read the problem and identify for which substance the mass is given and for which substance
the mass is required. Write a balanced chemical equation to determine the relative amounts of the
reactant and the product. Calculate their molar masses. Next, use the molar mass of the reactant
to convert its mass to its amount in moles. Then use the coefficients in the balanced chemical
equation to convert the amount of the reactant to the amount of the product. Finally, use the
molar mass of the product to convert its amount to its mass.
31. The air vents of a Bunsen burner permit the user to control the rate at which air, and thus
oxygen, combines with the gas in an air/fuel mixture. The user can close the vents to create a gas
mixture rich in fuel that is easy to ignite and then open the vents to allow more air to enter the
burner to produce a leaner, cleaner-burning mixture appropriate for heating glassware and other
materials in laboratory activities.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-3
32. Both baking soda and sodium hydroxide can neutralize acid spills. Using an excess amount
of a basic substance can ensure that the spill is completely consumed. Baking soda can be
applied in excess to acid spills because any baking soda remaining after the neutralization is
complete is not toxic or corrosive. If sodium hydroxide were applied in excess to an acid spill,
any sodium hydroxide remaining would pose a hazard, as sodium hydroxide is corrosive. It
would be inconvenient to attempt to apply a stoichiometric amount of sodium hydroxide to
prevent any from being left over.
33. In industrial chemical processes, the expensive reactants are usually made to be the limiting
reagents to ensure that none of these expensive chemicals is left over or wasted. If these reactants
were to be provided in excess, not all of them would be consumed. Unless the excess reagents
are recycled, they would become waste. Wasting expensive chemicals would be financially
unwise. Recycling could keep them from being wasted, but the recycling process would also add
unwanted cost to the process.
34. Chemical engineers have to ensure that no chemical needed for a manufacturing process runs
out. If this were to happen, the process would be halted and profit would be lost. At the same
time, the engineers must avoid maintaining an oversupply because this is costly and potentially
dangerous. By examining the relevant balanced chemical equations, the engineers can determine
which reactant is to be limiting (likely the most expensive one) and then use the mole ratios to
determine the amounts of the other chemicals that they have to keep in excess.
m
35. If the quantities of the reactants are not already given in moles, use n =
to convert them
M
to moles. Use one of the amounts and the mole ratio from the balanced chemical equation to
determine the stoichiometric amount of the other reactant. If the actual amount of this reactant is
greater than its stoichiometric amount, it is in excess. If the actual amount is less than the
stoichiometric amount, it is limiting.
36. In theory, an actual yield could be equal to the theoretical yield. However, experimental
errors could cause the actual yield to seem greater than the theoretical yield. For example, a
substance that is collected by filtration might still be damp when its mass is determined. The
moisture would add mass that would make the mass of product appear to be greater than it
really is.
37. When cisplatin is synthesized, a competing reaction produces transplatin, which lowers the
yield of cisplatin. Transplatin is not an effective cancer drug, so it cannot be sold. Thus, the
unavoidable “waste” of reactants in the formation of transplatin reduces profits. If the tranplatin
remains mixed with the cisplatin, it makes it difficult to measure out the correct dose.
Analysis and Application
38. (a) Convert amount of dinitrogen pentoxide to amounts of oxygen and nitrogen dioxide.
1 molO
4 molNO
2
2
nO = 0.50 molN O !
nNO = 0.50 molN O !
2
2 5
2
2 5
2 molN O
2 molN O
2
5
nO = 0.25 mol
2
2
5
nNO = 1.0 mol
2
Therefore, 0.50 mol of dinitrogen pentoxide will produce 0.25 mol of oxygen and 1.0 mol of
nitrogen dioxide.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-4
(b) Convert amount of nitrogen dioxide to amounts of dinitrogen pentoxide and oxygen.
2 molN O
1 molO
2 5
2
nN O = 1.60 molNO !
nO = 1.60 molNO !
2 5
2
2
2
4 molNO
4 molNO
2
2
nN O = 0.800 mol
2
nO = 0.400 mol
5
2
Therefore, 0.800 mol of dinitrogen pentoxide will produce 0.400 mol of oxygen and 1.60 mol of
nitrogen dioxide.
39. (a) Given: nC10H 22 = 5.50 mol
Required: amount of oxygen, nO2
Solution:
Step 1. List the given value and the required value.
2 C10H22(l) + 31 O2(g) → 20 CO2(g) + 22 H2O(g)
nO2
5.50 mol
Step 2. Convert amount of decane to amount of oxygen.
31 molO
2
nO = 5.50 molC H !
2
10 22
2 molC H
10
22
nO = 85.3 mol
2
Statement: Therefore, 85.3 mol of oxygen is required for the complete combustion of 5.50 mol
of decane.
(b) Given: nCO2 = 12 mol
Required: amount of water, n H 2O
Solution:
Step 1. List the given value and the required value.
2 C12H22(l) + 31 O2(g) → 20 CO2(g) + 22 H2O(g)
n H 2O
12 mol
Step 2. Convert amount of carbon dioxide to amount of water.
22 molH O
2
nH O = 12 molCO !
2
2
20 molCO
2
nH O = 13 mol
2
Statement: When 12 mol of carbon dioxide is produced, 13 mol of water is also produced.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: Stoichiometry in Chemical Reactions
7-5
40. Use the balanced chemical equations for the two reactions to determine the stoichiometric
mole ratio of copper to sulfur in each reaction.
2 Cu(s) + S(g) → Cu2S(s)
For the production of copper(I) sulfide, the mole ratio of copper to sulfur is 2:1.
Cu(s) + S(g) → CuS(s)
For the production of copper(II) sulfide, the mole ratio of copper to sulfur is 1:1.
The scientist could measure the initial mass of the copper wire and the mass of the copper sulfide
product. She could then subtract the mass of copper from the mass of product to get the mass of
sulfur that combined with the copper. Converting the masses of copper and sulfur to amounts
would provide the actual mole ratio in which the elements reacted. If the experimental mole ratio
of copper to sulfur was 2:1, the scientist’s hypothesis would be supported.
41. (a) 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
(b) Given: density of octane = 0.70 g/cm3; volume of octane = 10.0 L
Required: mass of carbon dioxide, mCO2
Solution:
Step 1. Convert volume of octane to mass of octane.
mC H = 10.0 L !
8
18
1000 cm 3 0.70 g
!
1L
1 cm 3
mC H = 7000 g
8
18
Step 2. List the given value, the required value, and the corresponding molar masses.
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
10.0 L
mCO2
7000 g
114.26 g/mol
44.01 g/mol
Step 3. Convert mass of octane to amount of octane.
1 molC H
8 18
nC H = 7000 g !
8 18
114.26 g
nC H = 61.264 mol [two extra digits carried]
8
18
Step 4. Convert amount of octane to amount of carbon dioxide.
16 mol
CO2
n = 61.264 molC H !
CO2
8 18
2 molC H
8
n
CO2
18
= 490.11 mol [two extra digits carried]
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Chapter 7: Stoichiometry in Chemical Reactions
7-6
Step 5. Convert amount of carbon dioxide to mass of carbon dioxide.
! 44.01 g $
mCO = (490.11 mol ) #
&
2
" 1 mol %
= 21 600 g
mCO = 22 kg
2
Statement: When 10.0 L of octane is burned in excess oxygen, 22 kg of carbon dioxide is
produced.
42. (a) In the reaction 2 H2(g) + O2(g) → 2 H2O(g), the stoichiometric mole ratio of oxygen to
hydrogen is 1:2.
(b) Given: ratio of mass of oxygen to mass of hydrogen = 4 g : 1 g
Required: mole ratio of oxygen to hydrogen
Solution:
Step 1. List the given values, the required values, and the corresponding molar masses.
2 H2(g) + O2(g) → 2 H2O(g)
1g
4g
nH2
nO2
2.02 g/mol 32.00 g/mol
Step 2. Convert the masses to amounts in moles.
1 molO
2
nO = 4 g !
2
32.00 g
nO = 0.125 mol [two extra digits carried]
2
nH = 1 g !
2
1 molH
2
2.02 g
nH = 0.495 mol [two extra digits carried]
2
Statement: The mole ratio of oxygen to hydrogen is 1:4.
(c) From the balanced equation in (a), 2 mol of hydrogen is required to react with 1 mol of
oxygen. From (b), the amount of hydrogen is greater than required. Therefore, hydrogen is in
excess in the fuel mixture with a ratio of 4 g O2 : 1 g H2.
43. Answers may vary. Sample answer: The relative numbers of parts used in building a tricycle
is analogous to the coefficients in a balanced chemical equation, which indicate the relative
amounts of reactants and products for a chemical reaction.
F + 3 W + 2 P → T (F = frame, W = wheel, P = pedal, T = tricycle)
Formation of Tricycles
Number of frames
3
12 dozen
6 mol
Number of wheels
9
36 dozen
18 mol
Copyright © 2011 Nelson Education Ltd.
Number of pedals
6
24 dozen
12 mol
Number of tricycles
3
12 dozen
6 mol
Chapter 7: Stoichiometry in Chemical Reactions
7-7
44. (a) 2 Cu(s) + O2(g) → 2 CuO(s)
(b) Given: amount of copper = 7.0 mol; amount of oxygen = 4.0 mol
Required: amount of copper(II) oxide produced, nCuO ; amount of excess reagent remaining
Solution:
Step 1. List the given values and the required value below the substances in the equation.
2 Cu(s) + O2(g) → 2 CuO(s)
7.0 mol 4.0 mol nCuO
Step 2. Determine the limiting reagent.
1 molO
2
nO = 7.0 molCu !
2
2 molCu
nO = 3.5 mol
2
Since the amount of oxygen present is greater than the required amount, copper is the limiting
reagent and oxygen is the excess reagent.
Step 3. Convert amount of copper to amount of copper(II) oxide.
2 molCuO
nCuO = 7.0 molCu !
2 molCu
nCuO = 7.0 mol
Step 4. Calculate the amount of excess oxygen remaining.
4.0 mol – 3.5 mol = 0.5 mol
Statement: The amount of copper(II) oxide produced is 7.0 mol and 0.5 mol of oxygen remains.
45. (a) Given: mC6H12O6 = 36.0 g
Required: mass of carbon, mC ; mass of water, m H 2O
Solution:
Step 1. List the given value, the required values, and the corresponding molar masses.
C6H12O6(s) → 6 C(g) + 6 H2O(l)
m H 2O
mC
3.60 g
180.18 g/mol
12.01 g/mol 18.02 g/mol
Step 2. Convert mass of glucose to amount of glucose.
1 molC H O
6 12 6
nC H O = 3.60 g !
6 12 6
180.18 g
nC H
6
12
O6
= 0.019 980 mol [two extra digits carried]
Step 3. Convert amount of glucose to amounts of carbon and water.
6 molC
nC = 0.019 980 molC H O !
6 12 6
1 molC H O
6
12
6
nC = 0.119 88 mol [two extra digits carried]
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Chapter 7: The Stoichiometry of Chemical Reactions
7-8
nH O = 0.019 980 molC H
2
6
12
O6
!
6 molH O
2
1 molC H
6
12
O6
nH O = 0.119 88 mol (two extra digits carried)
2
Step 4. Convert amounts of carbon and water to masses of carbon and water.
! 12.01 g $
mC = (0.119 88 mol ) #
&
" 1 mol %
mC = 1.44 g
! 18.02 g $
mH O = (0.119 88 mol ) #
&
2
" 1 mol %
mH O = 21.6 g
2
Statement: When 36.0 g of glucose decomposes completely into carbon and water, 1.44 g of
carbon and 21.6 g of water would be produced.
(b) Given: mC = 21.6 g ; m H 2O = 32.4 g
Required: mass of glucose decomposed, mC6H12O6
Solution:
Step 1. List the given value, the required values, and the corresponding molar masses.
C6H12O6(s) → 6 C(g) + 6 H2O(l)
mC6H12O6
21.6 g
32.4 g
180.18 g/mol
12.01 g/mol 18.02 g/mol
Step 2. Convert masses of carbon and water to amounts of carbon and water.
1 molC
nC = 21.6 g !
12.01 g
nC = 1.7985 mol [two extra digits carried]
nH O = 32.4 g !
2
1 molH O
2
18.02 g
nH O = 1.7980 mol [two extra digits carried]
2
Carbon and water are in stoichiometric amounts.
Step 3. Convert amount of carbon to amount of glucose.
1 molC H O
6 12 6
nC H O = 1.7980 molC !
6 12 6
6 molC
nC H
6
12
O6
= 0.29975 mol [two extra digits carried]
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Chapter 7: The Stoichiometry of Chemical Reactions
7-9
Step 4. Convert amount of glucose to mass of glucose.
! 180.18 g $
mC H O = (0.299 75 mol ) #
&
6 12 6
" 1 mol %
mC H
6
12
O6
= 54.0 g
Statement: If 21.6 g of carbon and 32.4 g of water are produced, 54.0 g of glucose must have
decomposed.
46. Given: n HgO = 0.500 mol
Required: mass of mercury, m Hg ; mass of oxygen, mO2
Solution:
Step 1. List the given value, the required values, and the corresponding molar masses.
2 HgO(s) → 2 Hg(l) + O2(g)
m Hg
mO 2
0.500 mol
216.59 g/mol 200.59 g/mol 32.00 g/mol
Step 2. Convert amount of mercury(II) oxide to amounts of mercury and oxygen.
2 molHg
nHg = 0.500 molHgO !
2 molHgO
nHg = 0.500 mol
nO = 0.500 molHgO !
2
1 molO
2
2 molHgO
nO = 0.250 mol
2
Step 3. Convert amounts of mercury and oxygen to masses of mercury and oxygen.
! 200.59 g $
mHg = (0.500 mol ) #
&
" 1 mol %
mHg = 100 g
! 32.00 g $
mO = (0.250 mol ) #
&
2
" 1 mol %
mO = 8.00 g
2
Statement: When 0.500 mol of mercury(II) oxide is fully decomposed, 100 g of mercury and
8.00 g of water are produced.
47. Answers may vary. Students’ answers will likely include appliances that burn fuels such as
furnaces, barbecue grills, gas ranges, and water heaters. In these instances, stoichiometry helps to
arrive at the correct air-to-fuel ratio. Other examples include glues or epoxy varnishes that
require the mixing of two components in proper proportions.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-10
48. (a) From the balanced chemical equation, 0.1 mol of aluminum requires 0.15 mol copper(II)
chloride for complete reaction. Therefore, the amount of copper(II) chloride is in excess. As a
result, the aluminum foil will be fully consumed, but some copper(II) chloride will remain. Thus,
the blue colour of the original solution will lighten, but will not disappear completely.
(b) In this case, the copper(II) chloride is the limiting reactant. The copper(II) chloride, therefore,
will be fully consumed instead of the aluminum. The blue colour of the original solution will
disappear completely, with some aluminum foil remaining in the solution.
49. Operators would have to provide an excess of calcium carbonate. If calcium carbonate were
limiting, not all of the sulfur dioxide present would be removed from the emissions.
50. (a) Given: mass of alloy = 10.00 g; mAgCl = 1.20 g
Required: mass of silver in original sample, mAg
Analysis: In the reaction of the silver-containing alloy in excess nitric acid,
2 Ag(s) + 2 HNO3(aq) → 2 AgNO3(aq) + H2(g), 1 mol of silver produces 1 mol of silver nitrate.
The amount of silver nitrate that reacts with sodium chloride equals the amount of silver
present in the original sample. The amount of silver chloride precipitated also equals the amount
of silver nitrate that reacts, and thus equals the amount of silver in the original sample.
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
n AgNO3
1.20 g
169.88 g/mol
143.32 g/mol
Step 2. Convert mass of silver chloride to amount of silver chloride.
1 molAgCl
nAgCl = 1.20 g !
143.32 g
nAgCl = 0.008 372 8 mol (two extra digits carried)
Step 3. Convert amount of silver chloride to amount of silver.
nAg = nAgNO
3
nAgNO = nAgCl
3
nAg = 0.008 372 8 mol (two extra digits carried)
Step 4. Convert amount of silver to mass of silver.
molar mass of silver = 107.87 g/mol
! 107.87 g $
mAg = (0.008 372 8 mol ) #
&
" 1 mol %
mAg = 0.903 18 g (two extra digits carried)
Statement: The mass of silver present in the original sample is 0.903 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-11
(b) mass of silver-containing alloy = 10.00 g
mass of silver present = 0.903 18 g
mass of silver
percentage of silver =
! 100 %
mass of alloy
0.903 18 g
=
! 100 %
10.00 g
percentage of silver = 9.03 %
The alloy contains 9.03 % silver by mass.
(c) Sodium chloride must be added in excess to ensure that all of the silver in the silver nitrate
solution is turned into a precipitate of silver chloride.
51. (a) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
(b) Given: density of methane = 0.70 kg/m3; volume of methane = 80 billion m3
Required: mass of carbon dioxide, mCO2
Solution:
Step 1. Convert volume of methane to mass in g.
mCH = 80 ! 109 m 3 !
4
0.70 kg
1 m3
!
1000 g
1 kg
mCH = 5.6 g ! 1013 g
4
Step 2. List the given value, the required value, and the corresponding molar masses.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
mCO2
5.6 × 1013 g
16.05 g/mol
44.01 g/mol
Step 3. Convert mass of methane to amount of methane.
1 molCH
4
nCH = 5.6 ! 1013 g !
4
16.05 g
nCH = 3.489 ! 1012 mol [two extra digits carried]
4
Step 4. Convert amount of methane to amount of carbon dioxide.
1 molCO
2
nCO = 3.489 ! 1012 molCH !
2
4
1 molCH
4
nCO = 3.489 ! 1012 mol [two extra digits carried]
2
Step 5. Convert amount of carbon dioxide to mass of carbon dioxide.
" 44.01 g %
mCO = (3.489 ! 1012 mol ) $
'
2
# 1 mol &
mCO = 1.536 ! 1014 g
2
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-12
Convert mass of carbon dioxide to mass in tonnes.
mCO = 1.5 ! 1014 g !
2
1 kg
1000 g
!
1 tonne
1000 kg
mCO = 1.5 ! 108 tonnes
2
Statement: The mass of carbon dioxide that is discharged into the atmosphere is
1.5 × 108 tonnes.
(c) Answers may vary. Sample answer: It is predicted that the carbon dioxide discharges into the
air would contribute to the Greenhouse effect. Gases in the layers of the atmosphere trap heat
and air particulates in the lower layers of the atmosphere that are close to Earth’s surface. The
trapped heat causes an increase in the surface temperature of Earth, causing climate changes,
such as warmer northern climates, where icecaps start melting at a faster rate than before. The
heat also causes changes in the tidal winds, which normally would move warm water to cold and
cold to warm, within a certain region. Rising water temperatures would force native species for
particular areas to move to colder waters. You will see a migration of species from traditional
waters to new waters.
52. The yellow-orange colour of the flame indicates incomplete combustion. Carbon particles
are, therefore, present in the flame in the form of soot because not all of the carbon in the
burning fuel is being converted to carbon dioxide gas. The deposit of soot on the bottom of the
pan causes blackening.
53. The balanced chemical equation for the combustion of natural gas indicates the
stoichiometric amounts of fuel and oxygen. The air/fuel mixture fed to the furnace could be
adjusted to match the stoichiometric amounts. This would make the combustion more efficient.
Efficient burning would, in turn, save money by keeping the amount of fuel consumed to a
minimum. Minimum consumption would also contribute to conservation of natural gas, a fossil
fuel. Moreover, efficient combustion would reduce air pollution caused by the undesirable or
“dirty” components produced in a fuel-rich flame, such as soot and carbon monoxide.
54. After the sulfur dioxide, SO2, residue is dissolved in water to form the sulfite ions, SO32–(aq),
a relatively large amount of hydrogen peroxide should be added to ensure that it is in excess and
will convert all of the sulfite ions to sulfate ions, SO42–(aq). Barium chloride, BaCl2(aq), should
then be added in small volumes with stirring. As long as each additional increment results in the
formation of precipitate, another small volume should be added. When no more precipitate
formation is observed, no more barium chloride should be added.
55. (a) Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
(b) In Trials 1 and 2, zinc is the limiting reagent and hydrochloric acid is the excess reagent
because increasing the quantity of zinc produced more hydrogen gas. In Trials 3 to 5,
hydrochloric acid is the limiting reagent and zinc is the excess reagent since increasing the
quantity of zinc from Trial 3 to Trial 4 and from Trial 4 to Trial 5 had no effect on the volume of
hydrogen gas collected in Trial 3.
56. Amounts in the Reaction of Sulfur Dioxide and Hydrogen Sulfide
Initial amount of
SO2(g)
(mol)
10.0
5.0
16.0
Initial amount of
H2S(g)
(mol)
30.0
8.0
34.0
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Amount of S8(s)
produced
(mol)
3.8
1.5
6.0
Amount of excess
reactant remaining
(mol)
10.0 H2S
1.0 SO2
2.0 H2S
Chapter 7: The Stoichiometry of Chemical Reactions
7-13
57. (a) Given: mAs2O3 = 19.8 g ; mZn = 32.7 g
Required: mass of arsine, mAsH 3
Solution:
Step 1. List the given values, the required value, and the corresponding molar masses.
As2O3(s) + 6 Zn(s) + 6 H2SO4(aq) → 2 AsH3(s) + 6 ZnSO4(s) + 3 H2O(l)
m AsH 3
19.8 g
32.7 g
197.84 g/mol 65.41 g/mol
77.95 g/mol
Step 2. Convert mass of given substances to amount of given substances.
1 molAs O
2 3
nAs O = 19.8 g !
2 3
197.84 g
nAs O = 0.100 01 mol [two extra digits carried]
2
3
nZn = 32.7 g !
1 molZn
65.41 g
nZn = 0.499 92 mol [two extra digits carried]
Step 3. Determine the limiting reagent.
6 molZn
nZn = 0.100 01 molAs O !
2 3
1 molAs O
2
3
nZn = 0.600 06 mol
Since the amount of zinc present is less than the required amount, zinc is the limiting reagent.
Step 4. Convert amount of zinc to amount of arsine.
2 molAsH
3
nAsH = 0.499 92 molZn !
3
6 molZn
nAsH = 0.166 64 mol [two extra digits carried]
3
Step 5. Convert amount of arsine to mass of arsine.
! 77.95 g $
mAsH = (0.166 64 mol ) #
&
3
" 1 mol %
mAsH = 13.0 g
3
Statement: The mass of AsH3 produced is 13.0 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-14
(b) Given: mass of As2O3 present = 19.8 g; nZn = 0.499 92 mol
Required: mass of As2O3 remaining
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
As2O3(s) + 6 Zn(s) + 6 H2SO4(aq) → 2 AsH3(s) + 6 ZnSO4(s) + 3 H2O(l)
n As2O3
0.499 92 mol
197.84 g/mol 65.41 g/mol
Step 2. Convert amount of zinc to amount of diarsenic trioxide, As2O3.
1 molAs O
2 3
nAs O = 0.499 92 molZn !
2 3
6 molZn
nAs O = 0.083 320 mol [two extra digits carried]
2
3
Step 3. Convert amount of As2O3 to mass of As2O3.
! 197.84 g $
mAs O = (0.083 320 mol ) #
&
2 3
" 1 mol %
mAs O = 16.484 g [two extra digits carried]
2
3
Step 4. Calculate the mass of As2O3 remaining.
19.8 g – 16.484 g = 3.32 g
Statement: The mass of As2O3 remaining is 3.32 g.
58. (a) Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
(b) Given: m Fe = 40.0 g ; m H 2SO4 = 100.0 g
Required: mass of hydrogen, m H 2
Solution:
Step 1. List the given values, the required value, and the corresponding molar masses.
Fe(s)
+ H2SO4(aq) → FeSO4(aq) + H2(g)
mH2
40.0 g
100.0 g
55.85 g/mol 98.09 g/mol
2.02 g/mol
Step 2. Convert mass of given substances to amount of given substances.
1 molFe
nFe = 40.0 g !
55.85 g
nFe = 0.716 20 mol [two extra digits carried]
nH SO = 100.0 g !
2
4
1 molH SO
2
4
98.09 g
nH SO = 1.019 47 mol [two extra digits carried]
2
4
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-15
Step 3. Determine the limiting reagent.
1 molH SO
2
4
nH SO = 0.716 20 molFe !
2
4
1 molFe
nH SO = 0.716 20 mol [two extra digits carried]
2
4
Since the amount of sulfuric acid present is greater than the required amount, iron is the limiting
reagent.
Step 4. Convert amount of iron to amount of hydrogen.
1 molH
2
nH = 0.716 20 molFe !
2
1 molFe
nH = 0.716 20 mol [two extra digits carried]
2
Step 5. Convert amount of hydrogen to mass of hydrogen.
! 2.02 g $
mH = (0.716 20 mol ) #
&
2
" 1 mol %
mH = 1.45 g
2
Statement: The mass of hydrogen produced is 1.45 g.
(c) Given: mass of sulfuric acid present = 100.0 g;
amount of sulfuric acid consumed, nH SO = 0.716 20 mol
2
4
Required: mass of sulfuric acid remaining
Solution:
Step 1. Convert amount of sulfuric acid consumed to mass of sulfuric acid.
molar mass of sulfuric acid = 98.09 g/mol
! 98.09 g $
mH SO = (0.716 20 mol ) #
&
2
4
" 1 mol %
mH SO = 70.252 g (two extra digits carried)
2
4
Step 2. Calculate the mass of sulfuric acid remaining.
100.0 g – 70.252 g = 29.8 g
Statement: The mass of sulfuric acid remaining is 29.8 g.
59. (a) Given: m NH 3 = 4500 kg ; mO2 = 7500 kg
Required: m NO
Solution:
Step 1. List the given values, the required values, and the corresponding molar masses.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
m NO
4500 kg
7500 kg
17.04 g/mol
32.00 g/mol 30.01 g/mol
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-16
Step 2. Convert mass of given substances to amount of given substances.
1000 g 1 molNH 3
nNH = 4500 kg !
!
3
17.01 g
1 kg
nNH = 264 600 mol [two extra digits carried]
3
nO = 7500 kg !
2
1000 g
1 kg
!
1 molO
2
32.00 g
nO = 234 400 mol [two extra digits carried]
2
Step 3. Determine the limiting reagent.
5 molO
2
nO = 264 600 molNH !
2
3
4 molNH
3
nO = 330 800 mol [two extra digits carried]
2
Since the amount of oxygen present is less than the required amount, oxygen is the limiting
reagent.
Step 4. Convert amount of oxygen to amount of nitrogen monoxide.
4 molNO
nNO = 234 400 molO !
2
5 molO
2
nNO = 187 500 mol (two extra digits carried)
Step 5. Convert amount of nitrogen monoxide to mass of nitrogen monoxide.
! 30.01 g $ ! 1 kg $
mNO = (187 500 mol ) #
&#
&
" 1 mol % " 1000 g %
mNO = 5600 kg
Statement: The mass of nitrogen monoxide that will form is 5600 kg.
(b) Given: mass of ammonia originally present = 4500 kg;
amount of oxygen consumed, nO2 = 234 400 mol
Required: mass of ammonia remaining
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
n NH 3
234 400 mol
17.04 g/mol 32.00 g/mol
Step 2. Convert amount of oxygen to amount of ammonia.
4 molO
2
nNH = 234 400 molNH !
3
3
5 molO
2
nNH = 187 500 mol [two extra digits carried]
3
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-17
Step 3. Convert amount of ammonia to mass of ammonia.
! 17.04 g $ ! 1 kg $
mNH = (187 500 mol ) #
&#
&
3
" 1 mol % " 1000 g %
mNH = 3195 kg [two extra digits carried]
3
Step 4. Calculate the mass of ammonia remaining.
4500 kg – 3195 kg = 1300 kg
Statement: The mass of ammonia remaining is 1300 kg.
60. Given: mAlP = 100.0 g ; m H 2O = 50.0 g
Required: m PH 3
Solution:
Step 1. List the given values, the required value, and the corresponding molar masses.
AlP(s) + 3 H2O(l) → PH3(g) + Al(OH)3(s)
m PH 3
100.0 g
50.0 g
57.95 g/mol 18.02 g/mol 34.00 g/mol
Step 2. Convert mass of given substances to amount of given substances.
1 molAlP
nAlP = 100.0 g !
57.95 g
nAlP = 1.725 63 mol [two extra digits carried]
nH O = 50.0 g !
2
1 molH O
2
18.02 g
nH O = 2.7747 mol [two extra digits carried]
2
Step 3. Determine the limiting reagent.
3 molH O
2
nH O = 1.725 63 molAlP !
2
1 molAlP
nH O = 5.1769 mol [two extra digits carried]
2
Since the amount of water present is less than the required amount, water is the limiting reagent.
Step 4. Convert amount of water to amount of phosphine.
1 molPH
3
nPH = 2.7747 molH O !
3
2
3 molH O
2
nPH = 0.924 90 mol [two extra digits carried]
3
Step 5. Convert amount of phosphine to mass of phosphine.
! 34.00 g $
mPH = (0.924 90 mol ) #
&
3
" 1 mol %
mPH = 31.4 g
3
Statement: The theoretical yield of phosphine is 31.4 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-18
61. Given: mAu = 3.00 kg ; percentage yield of NaAu(CN)2 = 95 %
Required: actual yield of NaAu(CN)2
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(l) → 4 NaAu(CN)2(aq) + 4 NaOH(aq)
m NaAu(CN)2
3.00 kg
196.97 g/mol
272.00 g/mol
Step 2. Convert mass of given substance to amount of given substance.
1000 g 1 molAu
nAu = 3.00 kg !
!
196.97 g
1 kg
nAu = 15.231 mol [two extra digits carried]
Step 3. Convert amount of gold to amount of NaAu(CN)2.
4 molNaAu(CN)
2
nNaAu(CN) = 15.231 molAu !
2
4 molAu
nNaAu(CN) = 15.231 mol [two extra digits carried]
2
Step 4. Convert amount of NaAu(CN)2to mass of NaAu(CN)2.
! 272.00 g $ ! 1 kg $
mNaAu(CN) = (15.231 mol ) #
&#
&
2
" 1 mol % " 1000 g %
mNaAu(CN) = 4.1428 kg [two extra digits carried]
2
This is the theoretical yield of NaAu(CN)2.
Step 5. Calculate the actual yield.
actual yield
percentage yield =
! 100 %
theoretical yield
actual yield = theoretical yield !
= 4.1428 kg !
percentage yield
100 %
95 %
100 %
actual yield = 3. 94 kg
Statement: The mass of NaAu(CN)2 extracted from 1.00 × 106 kg of ore is 3.94 kg.
62. (a) Given: mCH 3Cl = 20.0 g ; mC6H 6 = 25.0 g
Required: theoretical yield of toluene, mC6H5CH 3
Solution:
Step 1. List the given values, the required value, and the corresponding molar masses.
CH3Cl(g) + C6H6(l) → C6H5CH3(l) + HCl(g)
mC6H 5CH 3
20.0 g
25.0 g
50.49 g/mol
78.12 g/mol 92.15 g/mol
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-19
Step 2. Convert mass of given substances to amount of given substances.
1 molCH Cl
3
nCH Cl = 20.0 g !
3
50.49 g
nCH Cl = 0.396 12 mol [two extra digits carried]
3
nC H6 = 25.0 g !
6
1 molC H6
6
78.12 g
nC H6 = 0.320 02 mol [two extra digits carried]
6
Step 3. Determine the limiting reagent.
1 molC H6
6
nC H6 = 0.396 12 molCH Cl !
6
3
1 molCH Cl
3
nC H6 = 0.396 12 mol [two extra digits carried]
6
Since the amount of benzene present is less than the required amount, benzene is the limiting
reagent.
Step 4. Convert amount of benzene to amount of toluene.
1 molC H CH
6 5
3
nC H CH = 0.320 02 molC H !
6 5
3
6
6
1 molC H
6
6
nC H CH = 0.320 02 mol [two extra digits carried]
6
5
3
Step 5. Convert amount of toluene to mass of toluene.
! 92.15 g $
mC H CH3 = (0.320 02 mol ) #
&
6 5
" 1 mol %
mC H CH = 29.490 g [two extra digits carried]
6
5
Statement: The theoretical yield of toluene is 29.4 g.
(b) actual yield of toluene = 22.0 g
Use the theoretical yield of toluene from (a).
actual yield
percentage yield =
! 100 %
theoretical yield
22.0 g
=
! 100 %
29.490 g
percentage yield = 74.6 %
The percentage yield of toluene is 74.6 %.
(c) There may be competing side reactions that use up some of the reactants, reducing the actual
amount of toluene produced.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-20
63. (a) Given: actual yield of aspirin = 12.2 g; percentage yield of reaction = 72 %
Required: theoretical yield of aspirin
Solution:
actual yield
percentage yield =
! 100 %
theoretical yield
theoretical yield =
=
actual yield
! 100 %
percentage yield
12.2 g
! 100 %
72 %
theoretical yield = 16.94 g [two extra digits carried]
Statement: The theoretical yield of aspirin in the synthesis is 16.9 g.
(b) Given: mC9 H8O4 = 16.94 g
Required: mass of salicylic acid, mC7 H 6O3
Solution:
Step 1. List the given value, the required value, and the corresponding molar masses.
C7H6O3(s) + C4H6O3(l) → C9H8O4(s) + HC2H3O2(l)
m C 7 H 6O 3
16.94 g
138.13 g/mol
180.17 g/mol
Step 2. Convert mass of given substance to amount of given substance.
1 molC H O
9 8 4
nC H O = 16.94 g !
9 8 4
180.17 g
nC H O = 0.094 02 mol [two extra digits carried]
9
8
4
Step 3. Convert amount of aspirin to amount of salicylic acid.
1 molC H O
7 6 3
nC H O = 0.094 02 molC H O !
7 6 3
9 8 4
1 molC H O
9
8
4
nC H O = 0.094 02 mol [two extra digits carried]
7
6
3
Step 4. Convert amount of salicylic acid to mass of salicylic acid.
! 138.13 g $
mC H O = (0.094 02 mol ) #
&
7 6 3
" 1 mol %
mC H O = 13 g
7
6
3
Statement: The mass of salicylic acid present initially is 13 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-21
Evaluation
64. (a) Answers may vary. Sample answer: The design of the procedure is promising in some
ways. Carbon dioxide does combine with sodium hydroxide to form sodium carbonate, Na2CO3.
Thus, if the cup of sodium hydroxide pellets gains mass, the change in mass could be a direct
measure of the carbon dioxide given off by the crickets. However, without a control of the
experiment that has everything set up the same except the presence of crickets, the researcher
cannot be certain that a mass gain in the sodium hydroxide pellets is caused by carbon dioxide
from the crickets. Water vapour from the air, for example, could also react with sodium
hydroxide in the pellets, causing them to gain mass.
(b) CO2(g) + 2 NaOH(s) → Na2CO3(s) + 2 H2O(l)
65. Answers may vary. Sample answer: This is not a wise design. The amount of nitrogen in an
airbag that will properly decelerate an occupant of a moving car depends on the space between
the airbag and the occupant. Since there is more space between a front-seat passenger and the
passenger-side airbag than there is between the driver and the steering-wheel airbag, more
nitrogen, and therefore, more sodium azide is needed for the passenger’s airbag.
66. Answers may vary. Sample answer: The nitrogen and hydrogen that are supplied to the
reaction vessel are produced from natural gas, steam, and air. Energy is required to heat the
steam. The production of this energy likely involves burning fossil fuels and thus the release of
carbon dioxide, a greenhouse gas, into the environment. The machinery of the ammonia plant is
designed to recycle the nitrogen and hydrogen that leave the reaction vessel when ammonia is
drawn off. Recycling these gases means less energy and natural gas have to be used to keep a
steady supply of nitrogen and hydrogen flowing through the reaction vessel. Using less energy
means a lower operating cost for the plant owners and reduced emissions of carbon dioxide from
energy production. At the same time, natural gas, a valuable fossil fuel, is conserved.
67. Answers may vary. Sample answer: Measured increments of KIO3(aq) should be added with
stirring to the dissolved sample of KI(aq) until the colour from I2(aq) just disappears. From the
total amount of KIO3(aq) used to that point, we could determine the mass of KI(s) in the original
sample using stoichiometric calculations based on the overall reaction.
68. Answers may vary. Sample answer: The filter paper and precipitate could still be moist and
the moisture would add to the total mass, making it look as though more product was formed
than the theoretical quantity. Similarly, the precipitate may not have been rinsed thoroughly with
pure water during filtration to make sure that any chemicals dissolved in the liquid wetting the
precipitate and filter paper are washed off. The dried chemicals would add unwanted mass to the
dried filter paper and precipitate.
Reflect on Your Learning
69. Answers may vary. Sample answer: I realized that ratios can be used to establish relative
amounts or quantities of entities that can be applied to many specific situations. For example,
suppose it stated on a bag of fertilizer that 1 bag was sufficient to cover 150 m2 and I knew that
my family’s lawn covered 300 m2. I could calculate the number of fertilizer bags needed to cover
the lawn using the ratio of the number of fertilizer bags to the lawn area in square metres, m2,
which in this case is 1:150. For an area of 300 m2, I could use the proportion
1
150
=
2
300
to find
out that I would need 2 bags of fertilizer.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-22
70. Answers may vary. Students’ answers should include a specific concern and an action plan
with well-defined steps such as reading in the textbook, going to the library to read related
material in other textbooks, doing Internet research, forming study groups with peers, and
scheduling an appointment with the teacher.
71. Answers may vary. Students’ answers should include a plan that is logical and also realistic
in terms of time and resources needed.
Research
72. (a) The reaction that generates the nitrogen gas in sodium azide-propelled airbags is:
2 NaN3(s) → 2 Na(s) + 3 N2(g)
This reaction is the source of the sodium. The sodium can be removed by a reaction between
sodium and potassium nitrate:
10 Na(s) + 2 KNO3(s) → K2O(s) + 5 Na2O(s) + N2(g)
This reaction has the added benefit of producing more nitrogen for the inflation of the airbag.
The potassium and sodium oxides pose their own hazards, but are converted to an inert
silicate glass when they react with silicon dioxide that is present in the airbag. The reaction can
be shown as
K2O + Na2O + SiO2 → Na2K2SiO4 (glass)
(b) Passengers, emergency responders, and even tow-truck drivers would be exposed to sodium
if the proper sodium-removing reagents were not present in airbags.
(c) Sodium azide is a very toxic chemical and automotive engineers have designed other inflation
technologies that pose fewer chemical hazards. For example, one inflation system uses
compressed gas that is released into the airbag by a small chemical explosion that is set off when
an impact occurs.
73. Answers may vary. Students’ answers should detail how the structure of a non-stoichiometric
compound is related to its practical applications. Non-stoichiometric compounds have elemental
compositions that cannot be represented by small whole numbers. Some have very interesting
uses such as in semiconductors and superconductors. Yttrium barium copper oxide, YBa2Cu3O7,
is an important example. It was the first compound to become superconducting above the boiling
point of nitrogen.
74. Answers may vary. Students should create graphic organizers that present the comparison of
propane and butane as fuel gases in a manner that is logical and easy to interpret.
75. Answers may vary. The two main systems for delivery of proper air/fuel mixtures in internal
combustion engines are carburetors and fuel injection systems. Students’ diagrams should reveal
in some detail the parts of such systems, noting how the fuel delivery systems provide a proper
fuel mixture under many driving conditions.
76. Answers may vary. Students’ presentations should describe the chemistry of the production
of partially hydrogenated oils and the formation of trans fats as a type of competing reaction.
They should also describe alternatives to trans fats. Students’ visual aids should be well laid out,
legible, and logically organized.
Copyright © 2011 Nelson Education Ltd.
Chapter 7: The Stoichiometry of Chemical Reactions
7-23