Chapter 5–1
Chapter 5 Chemical Reactions
Solutions to In-Chapter Problems
5.1 The process is a chemical reaction because the reactants contain two gray spheres joined (indicating
H2) and two red spheres joined (indicating O2), while the product (H2 O) contains a red sphere joined
to two gray spheres (indicating O–H bonds).
5.2 The process is a physical change (freezing) since the particles in the reactants are the same as the
particles in the products.
5.3 Chemical equations are written with the reactants on the left and the products on the right
separated by a reaction arrow.
products
reactants
a.
2 H2O2(aq)
b.
2 C8H18
c.
5.4
+
2 H2O(l)
+
25 O2
2 Na3PO4(aq)
+
16 CO2
+
O2(g)
(4 H, 4 O)
18 H2O
(16 C, 50 O, 36 H)
Mg3(PO4)2(s) +
3 MgCl2(aq)
6 NaCl(aq)
(3 Mg, 2 P, 8 O, 6 Na, 6 Cl)
To determine the number of each type of atom when a formula has both a coefficient and a
subscript, multiply the coefficient by the subscript.
For 3 Al2(SO4)3: Al = 6 (3 × 2), S = 9 (3 × 3), O = 36 (3 × 3 × 4)
5.5
Write the chemical equation for the statement.
CH4(g) + 4 Cl2(g)
5.6
!
CCl4(l) + 4 HCl(g)
Balance the equation with coefficients one element at a time to have the same number of atoms
on each side of the equation. Follow the steps in Example 5.2.
[1] Place a 2 to balance O's.
a.
2 H2
+
O2
2 H2O
c.
CH4
+
2 Cl2
CH2Cl2
+
2 HCl
[2] Place a 2 to balance H's.
b.
5.7
2 NO
+ O2
Write the balanced chemical equation for carbon monoxide and oxygen reacting to form carbon
dioxide. The smallest set of whole numbers must be used.
2 CO + O2
5.8
2 NO2
2 CO2
Follow the steps in Example 5.2 to write the balanced chemical equation.
2 C2H6 + 7 O2
4 CO2 + 6 H2O
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–2
5.9
Write a balanced equation for the Haber process.
2 NH3
N2 + 3 H2
5.10
Balance the equations as in Example 5.2.
a.
b.
5.11
2 Al
+ 3 H2SO4
Al2(SO4)3
3 Na2SO3 + 2 H3PO4
+
3 H2SO3
3 H2
+
2 Na3PO4
One mole, abbreviated as mol, always contains an Avogadro’s number of particles (6.02 × 1023).
a, b, c, d: 6.02 × 1023
5.12
Multiply the number of moles by Avogadro’s number to determine the number of atoms.
Avogadro’s number is the conversion factor that relates moles to molecules, as in Example 5.3.
a. 2.00 mol × 6.02 × 1023 atoms/mol = 1.20 × 1024 atoms
b. 6.00 mol × 6.02 × 1023 atoms/mol = 3.61 × 1024 atoms
c. 0.500 mol × 6.02 × 1023 atoms/mol = 3.01 × 1023 atoms
d. 25.0 mol × 6.02 × 1023 atoms/mol = 1.51 × 1025 atoms
5.13
Multiply the number of moles by Avogadro’s number to determine the number of molecules, as
in Example 5.3.
a. 2.5 mol × 6.02 × 1023 molecules/mol = 1.5 × 1024 molecules
b. 0.25 mol × 6.02 × 1023 molecules/mol = 1.5 × 1023 molecules
c. 0.40 mol × 6.02 × 1023 molecules/mol = 2.4 × 1023 molecules
d. 55.3 mol × 6.02 × 1023 molecules/mol = 3.33 × 1025 molecules
5.14
Use Avogadro’s number as a conversion factor to relate molecules to moles.
a.
6.02 x 1025 molecules
x
b.
3.01 x 1022 molecules
x
c.
9.0 x 1024 molecules
1 mol
6.02 x 1023 molecules
1 mol
6.02 x 1023 molecules
x
1 mol
6.02 x 1023 molecules
= 100. mol
=
0.0500 mol
=
15 mol
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–3
5.15
To calculate the formula weight, multiply the number of atoms of each element by the atomic
weight and add the results.
a.
1 Ca atom
1 C atom
3 O atoms
40.08 amu
12.01 amu
16.00 amu
×
×
×
40.08 amu
12.01 amu
48.00 amu
=
=
=
Formula weight of CaCO3
b.
1 K atom
1 I atom
×
×
39.10 amu
126.9 amu
100.09 amu
39.10 amu
126.9 amu
=
=
Formula weight of KI
5.16
166.00 amu rounded to 166.0 amu
Calculate the molecular weight in two steps:
[1] Write the correct formula and determine the number of atoms of each element from the
subscripts.
[2] Multiply the number of atoms of each element by the atomic weight and add the results.
a.
b.
c.
2 C atoms
6 H atoms
1 O atom
12.01 amu
1.008 amu
16.00 amu
×
×
×
24.02 amu
6.048 amu
16.00 amu
=
=
=
Molecular weight of ethanol (C2H6 O)
46.068 amu rounded to 46.07 amu
6 H atoms
6 C atoms
1 O atom
6.048 amu
72.06 amu
16.00 amu
1.008 amu
12.01 amu
16.00 amu
×
×
×
=
=
=
Molecular weight of phenol (C6H6 O)
94.108 amu rounded to 94.11 amu
1 H atom
2 C atom
1 Br atom
1 Cl atom
3 F atoms
1.008 amu
24.02 amu
79.90 amu
35.45 amu
57.00 amu
1.008 amu
12.01 amu
79.90 amu
35.45 amu
19.00 amu
×
×
×
×
×
=
=
=
=
=
Molecular weight of halothane
(C2 HBrClF3)
197.378 amu rounded to 197.38 amu
5.17
C20H24O10
20 C atoms
24 H atoms
10 O atoms
×
×
×
12.01 amu
1.008 amu
16.00 amu
Molecular weight of ginkgolide B:
5.18
=
=
=
240.2 amu
24.192 amu
160.0
amu
424.392 amu = 424.4 g/mol
Convert the moles to grams using the molar mass as a conversion factor.
a. 0.500 mol of NaCl × 58.44 g/mol = 29.2 g
b. 2.00 mol of KI × 166.0 g/mol = 332 g
c. 3.60 mol of C2 H4 × 28.05 g/mol = 101 g
d. 0.820 mol of CH4 O × 32.04 g/mol = 26.3 g
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–4
5.19
Convert grams to moles using the molar mass as a conversion factor.
a.
1 mol
x
100. g NaCl
=
1.71 mol
=
1.59 mol
58.44 g
b.
25.5 g CH4
1 mol
x
16.04 g
c.
1 mol
x
0.250 g C9H8O4
1.39 x 10–3 mol
=
180.2 g
d.
5.20
25.0 g H2O
1 mol
x
=
1.39 mol
18.02 g
Use conversion factors to determine the number of molecules in 1.00 g; two 500.-mg tablets =
1.00 g.
1.00 g penicillin x
Answer:
6.02 x 1023 molecules
1.80 x 1021 molecules of penicillin
=
334.4 g penicillin
Grams cancel.
5.21
Use mole–mole conversion factors as in Example 5.5 and the equation below to solve the
problems.
N2(g)
+
!
O2(g)
2 NO(g)
a. 3.3 mol N2 × (2 mol NO/1 mol N2) = 6.6 mol NO
b. 0.50 mol O2 × (2 mol NO/1 mol O2) = 1.0 mol NO
c. 1.2 mol N2 × (1 mol O2/1 mol N2) = 1.2 mol O2
5.22
Use mole–mole conversion factors as in Example 5.5 and the equation below to solve the
problems.
2 C2H6(g)
+
5 O2(g)
!
4 CO(g)
+
6 H2O(g)
a. 3.0 mol C2 H6 × (5 mol O2/2 mol C2 H6) = 7.5 mol O2
b. 0.50 mol C2H6 × (6 mol H2 O/2 mol C2H6) = 1.5 mol H2 O
c. 3.0 mol CO × (2 mol C2 H6/4 mol CO) = 1.5 mol C2 H6
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–5
5.23
[1] Convert the number of moles of reactant to the number of moles of product using a mole–
mole conversion factor.
[2] Convert the number of moles of product to the number of grams of product using the
product’s molar mass.
2 C2H6O(aq)
C6H12O6(aq)
a.
+
2 CO2(g)
mole–mole
conversion factor
Moles of
reactant
2 mol C2H6O
x
0.55 mol C6H12O6
1 mol C6H12O6
Moles of
product
=
1.1 mol C2H6O
Moles C6H12O6 cancel.
1.1 mol C2H6O
Grams of
product
molar mass
conversion factor
Moles of
product
46.07 g C2H6O
x
51 g C2H6O
=
1 mol C2H6O
Answer
Moles cancel.
b.
mole–mole
conversion factor
Moles of
reactant
2 mol CO2
x
0.25 mol C6H12O6
Moles of
product
=
1 mol C6H12O6
0.50 mol CO2
Moles C6H12O6 cancel.
Moles of
product
0.50 mol CO2
molar mass
conversion factor
44.01 g CO2
x
Grams of
product
22 g CO2
=
1 mol CO2
Answer
Moles cancel.
c.
mole–mole
conversion factor
Moles of
product
x
1.0 mol C2H6O
Moles of
reactant
1 mol C6H12O6
=
2 mol C2H6O
0.50 mol C6H12O6
Moles C2H6O cancel.
molar mass
conversion factor
Moles of
reactant
0.50 mol C6H12O6
x
180.2 g C6H12O6
1 mol C6H12O6
Grams of
reactant
=
90. g C6H12O6
Answer
Moles cancel.
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–6
5.24
Use the steps outlined in Answer 5.23 to answer the questions.
C2H6O(l)
+
3 O2(g)
2 CO2(g)
+
3 H2O(g)
a. 0.50 mol C2 H6 O × (2 mol CO2/1 mol C2H6 O) = 1.0 mol CO2
1.0 mol CO2 × (44.01 g CO2/1 mol CO2) = 44 g CO2
b. 2.4 mol C2 H6O × (3 mol H2O/1 mol C2 H6O) = 7.2 mol H2 O
7.2 mol H2 O × (18.02 g H2O/1 mol H2 O) = 130 g H2 O
c. 0.25 mol C2 H6 O × (3 mol O2/1 mol C2 H6O) = 0.75 mol O2
0.75 mol O2 × (32.00 g O2/1 mol O2) = 24 g O2
5.25
Use conversion factors to solve the problems. Follow the steps in Sample Problem 5.14.
C7H6O3(s)
salicylic acid
+
C2H4O2(l)
acetic acid
C9H8O4(s)
aspirin
+ H2O(l)
a. 55.5 g C7 H6O3 × (1 mol C7H6 O3/138.1 g C7 H6O3) = 0.402 mol C7 H6O3
0.402 mol C7 H6O3 × (1 mol C9H8 O4/1 mol C7 H6O3) = 0.402 mol C9H8 O4
0.402 mol C9 H8O4 × (180.2 g C9H8 O4/1 mol C9 H8O4) = 72.4 g C9 H8O4
b. 55.5 g C7 H6 O3 × (1 mol C7 H6 O3/138.1 g C7H6 O3) = 0.402 mol C7 H6O3
0.402 mol C7 H6O3 × (1 mol C2H4 O2/1 mol C7 H6O3) = 0.402 mol C2H4 O2
0.402 mol C2 H4O2 × (60.05 g C2H4 O2/1 mol C2 H4O2) = 24.1 g C2 H4O2
c. 55.5 g C7 H6O3 × (1 mol C7H6 O3/138.1 g C7 H6O3) = 0.402 mol C7 H6O3
0.402 mol C7 H6O3 × (1 mol H2 O/1 mol C7 H6 O3) = 0.402 mol H2 O
0.402 mol H2 O × (18.02 g H2 O/1 mol H2 O) = 7.24 g H2O
5.26
Use conversion factors to solve the problems. Follow the steps in Sample Problem 5.14.
N2 + O2 → 2 NO
a. 10.0 g N2 × (1 mol N2/28.02 g N2) = 0.357 mol N2
0.357 mol N2 (2 mol NO/1 mol N2) = 0.714 mol NO
0.714 mol NO × (30.01 g NO/1 mol NO) = 21.4 g NO
b. 10.0 g O2 × (1 mol O2/32.00 g O2) = 0.313 mol O2
0.313 mol O2 × (2 mol NO/1 mol O2) = 0.626 mol NO
0.626 mol NO × (30.01 g NO/1 mol NO) = 18.8 g NO
c. 10.0 g N2 × (1 mol N2/28.02 g N2) = 0.357 mol N2
0.357 mol N2 × (1 mol O2/1 mol N2) = 0.357 mol O2
0.357 mol O2 × (32.00 g O2/1 mol O2) = 11.4 g O2
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–7
5.27
[1] Convert the number of moles of reactant to the number of moles of product using a mole–
mole conversion factor.
[2] Convert the number of moles of product to the number of grams of product—the theoretical
yield—using the product’s molar mass.
C(s)
+
O2(g)
CO2(g)
mole–mole
conversion factor
Moles of
reactant
1 mol CO2
x
3.50 mol C
Moles of
product
=
1 mol C
molar mass
conversion factor
Moles of
product
Grams of
product
44.01 g CO2
x
3.50 mol CO2
154 g CO2
=
1 mol CO2
Percent yield
actual yield (g)
=
theoretical yield (g)
=
5.28
3.50 mol CO2
53.5 g
154 g
x
Theoretical yield
Answer part (a)
x
100%
=
100%
34.7%
Answer part (b)
Use the steps in Answer 5.27 to solve the problem.
mole–mole
conversion factor
Moles of
reactant
2 mol O3
x
8.0 mol O2
Moles of
product
=
3 mol O2
molar mass
conversion factor
Moles of
product
x
5.3 mol O3
Grams of
product
48.00 g O3
=
1 mol O3
Percent yield
=
actual yield (g)
theoretical yield (g)
=
155 g
250 g
x
100%
5.3 mol O3
250 g O3
Theoretical yield
Answer part (a)
x
100%
=
62%
Answer part (b)
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–8
5.29
Use the steps in Sample Problem 5.17 to answer the questions.
H
H
C
HO
C
C
C
H
H
C
NH2
C
H
acetyl
chloride
4-aminophenol
molar mass 109.1 g/mol
a.
C
C
C
C
HO
C2H3ClO
+
H
C
C
H
H
=
mole–mole
conversion factor
Moles of
reactant
=
1 mol 4-aminophenol
0.733 mol acetaminophen x
Grams of
product
151.2 g acetaminophen
=
1 mol acetaminophen
=
HCl
0.733 mol acetaminophen
molar mass
conversion factor
=
+
Moles of
product
1 mol acetaminophen
x
Moles of
product
Percent yield
H
H
0.733 mol 4-aminophenol
109.1 g 4-aminophenol
b.
C
Moles of
reactant
1 mol 4-aminophenol
0.733 mol 4-aminophenol
C
H
molar mass
conversion factor
x
H
acetaminophen
molar mass 151.2 g/mol
Grams of
reactant
80.0 g 4-aminophenol
N
O
actual yield (g)
theoretical yield (g)
65.5 g
111 g
x
100%
x
111 g acetaminophen
Theoretical yield
100%
=
59.0%
Answer
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–9
5.30
Use conversion factors to solve the problem. Follow the steps in Answer 5.29.
a.
molar mass
conversion factor
Grams of
reactant
324 g O2
Moles of
reactant
1 mol O2
x
10.1 mol O2
=
32.00 g O2
mole–mole
conversion factor
Moles of
reactant
2 mol O3
x
10.1 mol O2
Moles of
product
=
Moles of
product
molar mass
conversion factor
6.73 mol O3
Grams of
product
48.00 g O3
x
Percent yield
=
=
5.31
actual yield (g)
theoretical yield (g)
122 g
323 g
x
100%
323 g O3
=
1 mol O3
b.
6.73 mol O3
3 mol O2
Theoretical yield
x
100%
=
37.8%
Answer
To determine the overall percent yield in a synthesis that has more than one step, multiply the
percent yield for each step.
a. (0.90)10 × 100% = 35%
b. (0.80)10 × 100% = 11%
c. 0.50 × (0.90)9 × 100% = 19%
d. 0.20 × 0.50 × 0.50 × 0.80 × 0.80 × 0.80 × 0.80 × 0.80 × 0.80 × 0.80 × 100% = 1.0%
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–10
5.32
Use the steps in Sample Problem 5.18 to answer the questions.
To determine the limiting reagent:
N
3 molecules of H2
? molecules of N2
original quantity
H
3 molecules H2
or
1 molecule N2
unknown quantity
1 molecule N2
Choose this conversion factor
to cancel molecules of H2.
3 molecules H2
3 molecules H2
1 molecule N2
x
=
3 molecules H2
1 molecule of N2 is needed.
2 molecules of N2 are left over.
2 molecules of NH3 are formed.
H2 is the limiting reactant.
N2
NH3
5.33
a.
b.
c.
d.
5.34
5.0 mol H2
1 mol O2
x
=
2.5 mol of O2 are needed.
Since 5.0 mol of O2 are present,
H2 is the limiting reactant.
2 mol H2
5.0 mol H2
x
8.0 mol H2
x
2.0 mol H2
x
1 mol O2
=
2.5 mol of O2 are needed.
Since 8.0 mol of O2 are present,
H2 is the limiting reactant.
=
4.0 mol of O2 are needed.
Since only 2.0 mol of O2 are present,
O2 is the limiting reactant.
=
1.0 mol of O2 is needed.
Since 5.0 mol of O2 are present, H2
is the limiting reactant.
2 mol H2
1 mol O2
2 mol H2
1 mol O2
2 mol H2
Calculate the number of moles of product formed as in Sample Problem 5.19.
mole–mole
conversion factor
a. 1.5 mol H2
x
1 mol N2
3 mol H2
=
0.50 mol N2 needed
H2 is the limiting reactant.
mole–mole
conversion factor
1.5 mol H2
x
2 mol NH3
3 mol H2
=
1.0 mol NH3
Answer
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–11
mole–mole
conversion factor
b. 1.0 mol H2
1 mol N2
x
=
3 mol H2
0.33 mol N2 needed
H2 is the limiting reactant.
mole–mole
conversion factor
1.0 mol H2
2 mol NH3
x
3 mol H2
=
0.67 mol NH3
Answer
mole–mole
conversion factor
c. 2.0 mol H2
1 mol N2
x
=
3 mol H2
0.67 mol N2 needed
H2 is the limiting reactant.
mole–mole
conversion factor
2.0 mol H2
2 mol NH3
x
3 mol H2
=
1.3 mol NH3
Answer
mole–mole
conversion factor
d. 7.5 mol H2
1 mol N2
x
=
3 mol H2
2.5 mol N2 needed
N2 is the limiting reactant.
mole–mole
conversion factor
2.0 mol N2
5.35
2 mol NH3
x
1 mol N2
=
4.0 mol NH3
Answer
Convert the number of grams of each reactant to the number of moles using molar masses. Since
the mole ratio of O2 to N2 is 1:1, the limiting reactant has fewer moles.
a.
molar mass
conversion factor
Grams of
reactant
12.5 g N2
x
1 mol N2
Moles of
reactant
=
28.02 g N2
molar mass
conversion factor
Grams of
reactant
15.0 g O2
x
1 mol O2
32.00 g O2
0.446 mol N2
N2 is the limiting reactant.
Moles of
reactant
=
0.469 mol O2
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–12
b.
molar mass
conversion factor
Grams of
reactant
14.0 g N2
Moles of
reactant
1 mol N2
x
0.500 mol N2
=
28.02 g N2
molar mass
conversion factor
Grams of
reactant
13.0 g O2
Moles of
reactant
1 mol O2
x
0.406 mol O2
=
32.00 g O2
5.36
O2 is the limiting reactant.
Calculate the number of moles of product formed based on the limiting reactant. Then convert
moles to grams using molar mass.
a.
0.446 mol N2
x
2 mol NO
=
0.892 mol NO
=
26.8 g NO
=
0.812 mol NO
=
24.4 g NO
1 mol N2
x
0.892 mol NO
30.01 g
1 mol NO
b.
0.406 mol O2
x
2 mol NO
1 mol O2
x
0.812 mol NO
30.01 g
1 mol NO
5.37
molar mass
conversion factor
Grams of
reactant
5.00 g H2
x
10.0 g O2
x
0.313 mol O2
Moles of
reactant
1 mol H2
=
2.48 mol H2
=
0.313 mol O2
2.016 g H2
1 mol O2
32.00 g O2
x
2 mol H2O
=
0.626 mol H2O
=
11.3 g H2O
O2 is the limiting reactant.
1 mol O2
0.626 mol H2O
x
18.02 g
1 mol H2O
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–13
5.38
A compound that gains electrons is reduced.
A compound that loses electrons is oxidized.
(oxidized)
a. Zn(s) +
(reduced)
2 H+(aq)
Zn2+
Zn
2 H+
b.
(reduced)
Fe3+(aq)
+
(oxidized)
Al(s)
+
3e
d.
Fe
2 Ag+
2 Ag
+
Br2
2 Ag+ (reduced)
Br2
+
2 Br–
2 Br–
2 AgBr
2 Br–
–
+
2 e–
+
2 e–
2 Br– (oxidized)
3 e–
+
+
Br2
Al3+(aq) + Fe(s)
I2
I2
2 I–
H2
Al3+
Fe3+
(oxidized) (reduced)
c. 2 I– + Br2
2 e–
+
2 e–
+
Al
5.39
Zn2+(aq) + H2(g)
2 e–
+
2 e–
2 Ag
A compound that gains electrons while causing another compound to be oxidized is called an
oxidizing agent.
A compound that loses electrons while causing another compound to be reduced is called a
reducing agent.
a. Zn reducing agent, H+ oxidizing agent
b. Fe3+ oxidizing agent, Al reducing agent
c. I– reducing agent, Br2 oxidizing agent
d. Br– reducing agent, Ag+ oxidizing agent
5.40
Zn is oxidized, and Hg2+ is reduced.
Zn2+
Zn
2 e–
+
Hg2+
2 e–
Hg
Hg2+ gains electrons and is reduced.
Zn loses electrons and is oxidized.
5.41
+
H2 is oxidized since it gains an O atom and C2H4 O2 is reduced since it gains hydrogen.
gains H atoms
reduced
C2H4O2
+
C2H6O
2 H2
+
H2O
gains O atom
oxidized
5.42
Zn is the reducing agent and Hg2+ is the oxidizing agent.
reduced
Zn
+
Hg2+
Zn2+
+
Hg
oxidized
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–14
Solutions to End-of-Chapter Problems
5.43 The process is a chemical reaction because the spheres in the reactants are joined differently than
the spheres in the products.
2 CO + 2 O3
2 CO2 + 2 O2
(not balanced)
5.44 a. The transformation of [1] to [2] is a chemical reaction because the spheres in the reactants (AB)
are joined differently than the spheres in the products (A2 and B2).
b. The transformation of [1] to [3] is a physical change because the spheres are joined the same
(AB) but they are now closer together indicating a physical state change from gas to liquid.
5.45 The difference between a coefficient and a subscript is that the coefficient indicates the number of
molecules or moles undergoing reaction, whereas the subscript indicates the number of atoms of
each element in a chemical formula.
5.46 It is not possible to change the subscripts of a chemical formula to balance an equation because
changing the subscripts changes the identity of the compound.
5.47 Add up the number of atoms on each side of the equation and then label the equations as balanced
or not balanced.
+
a. 2 HCl(aq)
CaCl2(aq)
Ca(s)
+
H2(g)
+
HCl
2 H, 2 Cl, 1 Ca: both sides, therefore balanced
b.
+
TiCl4
TiO2
2 H2O
1 Ti, 1 Cl, 1 H, 2 O
1 Ti, 4 Cl, 4 H, 2 O
NOT balanced
c.
+
Al(OH)3
AlPO4
H3PO4
+
3 H2O
1 Al, 1 P, 7 O, 6 H: both sides, therefore balanced
5.48 Add up the number of atoms on each side of the equation and then label the equations as balanced
or not balanced.
+
a. 3 NO2
H2O
+
HNO3
3 N, 7 O, 2H
2 NO
3 N, 5 O, 1 H
NOT balanced
+
b. 2 H2S
3 O2
+
H2O
2 SO2
2 H, 2 S, 5 O
4 H, 2 S, 6 O
NOT balanced
+
c. Ca(OH)2
2 H2O
2 HNO3
+
Ca(NO3)2
1 Ca 8 O, 4 H, 2 N: both sides, therefore balanced
5.49 Write the balanced equation using the colors of the spheres to identify the atoms (gray = hydrogen
and green = chlorine).
H2
+
Cl2
2 HCl
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–15
5.50 Write the balanced equation using the colors of the spheres to identify the atoms (red = oxygen and
blue = nitrogen).
O2
+
NO
NO3
5.51 Balance the equation with coefficients one element at a time to have the same number of atoms on
each side of the equation. Follow the steps in Example 5.2.
a.
Ni(s)
+
2 HCl(aq)
NiCl2(aq)
+
H2(g)
b.
CH4(g)
+
4 Cl2(g)
CCl4(g)
+
4 HCl(g)
c.
2 KClO3
d.
Al2O3
+
3 H2O
e.
4 Al(OH)3 + 6 H2SO4
2 KCl
+
6 HCl
+
3 O2
2 AlCl3
2 Al2(SO4)3 +
12 H2O
5.52 Balance the equation with coefficients one element at a time so that there are the same numbers of
atoms on each side of the equation. Follow the steps in Example 5.2.
a.
Mg(s)
+
b.
2 CO(g) +
c.
2 PbS(s) + 3 O2(g)
d.
H2SO4
e.
2 H3PO4 + 3 Ca(OH)2
2 HBr(aq)
+
O2(g)
2 NaOH
MgBr2(aq) +
H2(g)
2 CO2(g)
2 PbO(s)
+
2 SO2(g)
Na2SO4
+
2 H2O
Ca3(PO4)2 +
6 H2O
5.53 Follow the steps in Example 5.2 and balance the equations.
a. 2 C6H6 + 15 O2
12 CO2 + 6 H2O
b. C7H8 + 9 O2
7 CO2 + 4 H2O
c. 2 C8H18 + 25 O2
16 CO2 + 18 H2O
5.54 Follow the steps in Example 5.2 and balance the equation.
2 C5H12O + 15 O2
10 CO2 + 12 H2O
5.55 Follow the steps in Example 5.2 and balance the equation.
2 S(s) + 3 O2(g) + 2 H2O(l)
2 H2SO4(l)
5.56 Follow the steps in Example 5.2 and balance the equation.
MgCl2 + 2 NaOH
Mg(OH)2 + 2 NaCl
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–16
5.57 Fill in the molecules of the products using the balanced equation and following the law of
conservation of mass. Each side must have the same number of O and C atoms.
O2
O3
CO2
CO
reactants
products
5.58 Fill in the molecules of the products using the balanced equation and following the law of
conservation of mass. Each side must have the same number of C, O, and N atoms.
N2
CO
CO2
NO
reactants
products
5.59 To calculate the formula weight, multiply the number of atoms of each element by the atomic
weight and add the results. The formula weight in amu is equal to the molar mass in g/mol.
a.
1 Na atom
1 N atom
2 O atoms
×
×
×
22.99 amu
14.01 amu
16.00 amu
=
=
=
Formula weight of NaNO2
b.
2 Al atom
3 S atoms
12 O atoms
×
×
×
26.98 amu
32.07 amu
16.00 amu
69.00 amu = 69.00 g/mol
=
=
=
Formula weight of Al2(SO4)3
c.
6 C atom
8 H atoms
6 O atoms
×
×
×
12.01 amu
1.008 amu
16.00 amu
Formula weight of C6 H8O6
22.99 amu
14.01 amu
32.00 amu
53.96 amu
96.21 amu
192.0 amu
342.17 amu rounded to 342.2 amu = 342.2 g/mol
=
=
=
72.06 amu
8.064 amu
96.00 amu
176.124 amu rounded to 176.12 amu = 176.12 g/mol
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–17
5.60 To calculate the formula weight, multiply the number of atoms of each element by the atomic
weight and add the results. The formula weight in amu is equal to the molar mass in g/mol.
a.
1 Mg atom
1 S atom
4 O atoms
×
×
×
24.30 amu
32.07 amu
16.00 amu
=
=
=
Formula weight of MgSO4
b.
3 Ca atoms
2 P atoms
8 O atoms
×
×
×
40.08 amu
30.97 amu
16.00 amu
120.37 amu = 120.37 g/mol
=
=
=
Formula weight of Ca3(PO4)2
c.
16 C atoms
16 H atoms
1 Cl atom
1 N atom
2 O atoms
2 S atoms
×
×
×
×
×
×
12.01 amu
1.01 amu
35.45 amu
14.01 amu
16.00 amu
32.07 amu
24.30 amu
32.07 amu
64.00 amu
120.24 amu
61.94 amu
128.00 amu
310.18 amu = 310.18 g/mol
=
=
=
=
=
=
Formula weight of C16H16ClNO2S
5.61
192.16 amu
16.16 amu
35.45 amu
14.01 amu
32.00 amu
64.14 amu
353.92 amu = 353.92 g/mol
Determine the molecular formula of L-dopa. Then calculate the formula weight and molar mass
as in Answer 5.59.
HO
H
C
C
C
HO
C
C
C
H
H
H
C
C
H
NH2 OH
O
C
a. molecular formula = C9H11NO4!!
b. formula weight = 197.2 amu!
c. molar mass = 197.2 g/mol
H
L-dopa
5.62 Determine the molecular formula of niacin. Then calculate the formula weight and molar mass as
in Answer 5.59
H
H
C
H
O
C
C
C
N
C
OH
C
a. molecular formula = C6H5NO2!!
b. formula weight = 123.1 amu!
c. molar mass = 123.1 g/mol
H
niacin
5.63 Convert all of the units to moles, and then compare the atomic mass or formula weight to determine
the quantity with the larger mass.
a. 1 mol of Fe atoms (55.85 g/mol) < 1 mol of Sn atoms (118.7 g/mol)
b. 1 mol of C atoms (12.01 g/mol) < 6.02 × 1023 N atoms = 1 mol N atoms (14.01 g/mol)
c. 1 mol of N atoms (14.01 g/mol) < 1 mol of N2 molecules = 2 mol N atoms (28.02 g/mol N2)
d. 1 mol of CO2 molecules (44.01 g/mol) > 3.01 × 1023 N2 O molecules = 0.500 mol N2 O (44.02
g/mol N2 O) = 22.01 g N2 O
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–18
5.64 Convert all of the units to moles, and then compare the atomic mass or formula weight to determine
the quantity with the larger mass.
a. 1 mol of Si atoms (28.08 g/mol) < 1 mol of Ar atoms (39.95 g/mol)
b. 1 mol of He atoms (4.00 g/mol) > 6.02 × 1023 H atoms = 1 mol H atoms (1.01 g/mol)
c. 1 mol of Cl atoms (35.45 g/mol) < 1 mol of Cl2 molecules = 2 mol Cl atoms (70.90 g/mol Cl2)
d. 1 mol of C2 H4 molecules (28.06 g/mol) > 3.01 × 1023 C2 H4 molecules = 0.500 mol C2 H4 (28.06
g/mol C2 H4) = 14.03 g C2H4
5.65 Calculate the molar mass of each compound as in Answer 5.59, and then multiply by 5.00 mol.
a. HCl = 182 g
b. Na2SO4 = 710. g
c. C2H2 = 130. g
d. Al(OH)3 = 390. g
5.66 Calculate the molar mass of each compound as in Answer 5.59, and then multiply by 0.50 mol.
a. NaOH = 20. g
b. CaSO4 = 68 g
c. C3H6 = 21 g
d. Mg(OH)2 = 29 g
5.67 Convert the grams to moles using the molar mass as a conversion factor.
0.500 g
a.
x
1 mol
=
1.46 x 10–3 mol
c.
25.0 g
5.00 g
x
1 mol
=
0.0730 mol
=
7.30 x 10–5 mol
=
0.139 mol
=
1.39 x 10–4 mol
342.3 g
342.3 g
b.
1 mol
x
=
0.0146 mol
d.
1 mol
0.0250 g x
342.3 g
342.3 g
5.68 Convert the grams to moles using the molar mass as a conversion factor.
0.500 g
a.
x
1 mol
=
2.77x 10–3 mol
c.
25.0 g
x
180.2 g
180.2 g
b.
5.00 g
x
1 mol
180.2 g
1 mol
=
0.0277 mol
d.
0.0250 g x
1 mol
180.2 g
5.69 Multiply the number of moles by Avogadro’s number to determine the number of molecules, as in
Example 5.3.
a. 2.00 mol × 6.02 × 1023 molecules/mol = 1.20 × 1024 molecules
b. 0.250 mol × 6.02 × 1023 molecules/mol = 1.51 × 1023 molecules
c. 26.5 mol × 6.02 × 1023 molecules/mol = 1.60 × 1025 molecules
d. 222 mol × 6.02 × 1023 molecules/mol = 1.34 × 1026 molecules
e. 5.00 × 105 mol × 6.02 × 1023 molecules/mol = 3.01 × 1029 molecules
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–19
5.70 Use Avogadro’s number to convert the number of molecules to moles.
1 mol
a. 5.00 x 1019 molecules x
23
6.02 x 10
= 8.31 x 10–5 g
molecules
1 mol
b. 6.51 x 1028 molecules x
23
6.02 x 10
= 1.08 x 105 g
molecules
1 mol
c. 8.32 x 1021 molecules x
6.02 x 10
23
= 1.38 x 10-2 g
molecules
1 mol
d. 3.10 x 1020 molecules x
6.02 x
1023
= 5.15 x 10-4 g
molecules
5.71 Use the molar mass as a conversion factor to convert the moles to grams. Use Avogadro’s number
to convert the number of molecules to moles.
a.
b.
3.60 mol
x
0.580 mol x
90.08 g
1 mol
90.08 g
1 mol
c.
7.3 x 1024 molecules x
d.
6.56 x 1022 molecules x
=
324 g
=
52.2 g
1 mol
6.02 x
1023
molecules
1 mol
6.02 x
1023
molecules
x
x
90.08 g
1 mol
90.08 g
1 mol
= 1.1 x 103 g
= 9.82 g
5.72 Use the molar mass as a conversion factor to convert the moles to grams. Use Avogadro’s number
to convert the number of molecules to moles.
a.
b.
3.6 mol
0.58 mol
x
x
384.7 g
1 mol
384.7g
1 mol
c.
7.3 x 1024 molecules x
d.
6.56 x 1022 molecules x
1.4 x 103 g
=
=
2.2 x 102 g
1 mol
6.02 x
1023
molecules
1 mol
6.02 x
1023
molecules
x
x
384.7 g
1 mol
384.7 g
1 mol
= 4.7 x 103 g
= 41.9 g
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–20
5.73
2 H C C H
+
+
4 CO2
5 O2
2 H2O
acetylene
a. 12.5 moles of O2 are needed to react completely with 5.00 mol of C2 H2.
5.00 mol C2 H2 × (5 mol O2/2 mol C2H2) = 12.5 mol O2
b. 12 moles of CO2 are formed from 6.0 mol of C2 H2.
6.0 mol C2 H2 × (4 mol CO2/2 mol C2H2) = 12 mol CO2
c. 0.50 moles of H2 O are formed from 0.50 mol of C2 H2.
0.50 mol C2 H2 × (2 mol H2 O/2 mol C2 H2) = 0.50 mol H2 O
d. 0.40 moles of C2H2 are needed to form 0.80 mol of CO2.
0.80 mol CO2 × (2 mol C2 H2/4 mol CO2) = 0.40 mol C2H2
5.74
2 Na(s) +
2 NaOH(aq)
2 H2O(l)
+
H2(g)
a. 3.0 moles of H2O are needed to react completely with 3.0 mol of Na.
3.0 mol Na × (2 mol H2 O/2 mol Na) = 3.0 mol H2 O
b. 0.19 moles of H2 are formed from 0.38 mol of Na.
0.38 mol Na × (1 mol H2/2 mol Na) = 0.19 mol H2
c. 1.82 moles of H2 are formed from 3.64 mol of H2 O.
3.64 mol H2 O × (1 mol H2/2 mol H2 O) = 1.82 mol H2
5.75 Use conversion factors as in Example 5.6 to solve the problems.
a. 220 g of CO2 are formed from 2.5 mol of C2H2.
b. 44 g of CO2 are formed from 0.50 mol of C2 H2.
c. 4.5 g of H2 O are formed from 0.25 mol of C2H2.
d. 240 g of O2 are needed to react with 3.0 mol of C2 H2.
5.76 Use conversion factors as in Example 5.6 to solve the problems.
a. 120 g of NaOH are formed from 3.0 mol of Na.
b. 0.30 g of H2 are formed from 0.30 mol of Na.
c. 3.6 g of H2 O are needed to react with 0.20 mol of Na.
5.77 Use the equation to determine the percent yield.
Percent yield
=
=
actual yield (g)
theoretical yield (g)
9.0 g
12.0 g
x 100%
x 100%
=
75%
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–21
5.78 Use the equation to determine the percent yield.
Percent yield
actual yield (g)
=
theoretical yield (g)
17.0 g
=
x 100%
x 100%
20.0 g
=
85.0 %
5.79 Use the following equations to determine the percent yield.
a.
molar mass
conversion factor
Grams of
reactant
x
3.20 g CH4
Moles of
reactant
1 mol CH4
=
16.04 g CH4
mole–mole
conversion factor
Moles of
reactant
0.200 mol CH4
x
Moles of
product
1 mol CHCl3
=
1 mol CH4
0.200 mol CHCl3
molar mass
conversion factor
Moles of
product
0.200 mol CHCl3
x
119.4 g CHCl3
Grams of
product
=
=
actual yield (g)
23.9 g CHCl3
Theoretical yield
x 100%
theoretical yield (g)
15.0 g CHCl3
23.9 g CHCl3
=
1 mol CHCl3
b. Percent yield
0.200 mol CH4
x
100%
=
62.8%
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–22
5.80
Use the following equations to determine the percent yield.
a.
molar mass
conversion factor
Grams of
reactant
x
48.0 g CH4O
Moles of
reactant
1 mol CH4O
=
32.0 g CH4O
mole–mole
conversion factor
Moles of
reactant
1.50 mol CH4O
x
Moles of
product
2 mol CO2
=
2 mol CH4O
1.50 mol CO2
molar mass
conversion factor
Moles of
product
x
1.50 mol CO2
44.0 g CO2
Grams of
product
66.0 g CO2
=
1 mol CO2
b. Percent yield
1.50 mol CH4O
=
actual yield (g)
theoretical yield (g)
48.0 g CO2
=
66.0 g CO2
Theoretical yield
x 100%
x 100%
=
72.7%
5.81
a.
b.
c.
4 molecules A
4 molecules A
4 molecules A
x
x
x
1 molecule B
1 molecule A
1 molecule B
2 molecules A
2 molecules B
1 molecule A
=
4 molecules of B are needed.
Molecule A is in excess.
Molecule B is the limiting reactant.
=
2 molecules of B are needed.
Molecule B is in excess.
Molecule A is the limiting reactant.
=
8 molecules of B are needed.
Molecule A is in excess.
Molecule B is the limiting reactant.
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–23
5.82
4 molecules A2
1 molecule B2
x
=
2 molecules A2
2 molecules of B2 are needed.
Molecule A2 is in excess.
Molecule B2 is the limiting reactant.
A2
A2
A2B
B2
reactants
products
5.83
a.
b.
1.0 mol NO
2.0 mol NO
x
x
1 mol O2
2 mol NO
1 mol O2
=
0.50 mol of O2 is needed.
O2 is in excess.
NO is the limiting reactant.
=
1.0 mol of O2 is needed.
NO is in excess.
O2 is the limiting reactant.
=
0.333 mol NO
=
0.313 mol O2
=
0.167 mol of O2 is needed.
O2 is in excess.
NO is the limiting reactant.
=
0.933 mol NO
=
0.500 mol O2
=
0.467 mol of O2 is needed.
O2 is in excess.
NO is the limiting reactant.
2 mol NO
c.
d.
10.0 g NO
x
10.0 g O2
x
0.333 mol NO
x
28.0 g NO
x
16.0 g O2
x
0.933 mol NO
x
1 mol NO
30.01 g NO
1 mol O2
32.00 g O2
1 mol O2
2 mol NO
1 mol NO
30.01 g NO
1 mol O2
32.00 g O2
1 mol O2
2 mol NO
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–24
5.84
a.
2.0 mol NO
1 mol O2
x
=
1.0 mol of O2 is needed.
O2 is in excess.
NO is the limiting reactant.
=
2.1 mol of O2 is needed.
NO is in excess.
O2 is the limiting reactant.
=
0.500 mol NO
=
0.313 mol O2
=
0.250 mol of O2 is needed.
O2 is in excess.
NO is the limiting reactant.
=
0.333 mol NO
=
0.125 mol O2
=
0.166 mol of O2 is needed.
NO is in excess.
O2 is the limiting reactant.
2 mol NO
b.
4.2 mol NO
1 mol O2
x
2 mol NO
c.
15.0 g NO
1 mol NO
x
30.01 g NO
10.0 g O2
x
0.500 mol NO
x
1 mol O2
32.00 g O2
1 mol O2
2 mol NO
d.
10.0 g NO
1 mol NO
x
30.01 g NO
4.0 g O2
x
0.333 mol NO
x
1 mol O2
32.00 g O2
1 mol O2
2 mol NO
5.85 Use the limiting reactant from Problem 5.83 to determine the amount of product formed. The
conversion of moles of limiting reagent to grams of product is combined in a single step.
a.
b.
c.
d.
2 mol NO2
1.0 mol NO x
2 mol NO
2 mol NO2
0.50 mol O2 x
0.333 mol NO
0.933 mol NO
1 mol O2
x
x
2 mol NO2
2 mol NO
2 mol NO2
2 mol NO
46.01 g NO2
x
1 mol NO2
46.01 g NO2
x
1 mol NO2
x
x
46.01 g NO2
1 mol NO2
46.01 g NO2
1 mol NO2
=
46 g NO2
=
46 g NO2
=
15.3 g NO2
=
42.9 g NO2
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–25
5.86 Use the limiting reactant from Problem 5.84 to determine the amount of product formed. The
conversion of moles of limiting reagent to grams of product is combined in a single step.
a.
2.0 mol NO x
b.
2.0 mol O2 x
46.01 g NO2
x
2 mol NO
c. 0.500 mol NO x
d.
2 mol NO2
2 mol NO2
46.01 g NO2
x
1 mol O2
2 mol NO2
1 mol NO2
46.01 g NO2
x
2 mol NO
92 g NO2
=
180 g NO2
=
23.0 g NO2
1 mol NO2
2 mol NO2
0.125 mol O2 x
=
1 mol NO2
x
1 mol O2
46.01 g NO2
=
1 mol NO2
11.5 g NO2
5.87
a.
8.00 g C2H4
x
12.0 g HCl
x
1 mol C2H4
=
0.285 mol C2H4
=
0.329 mol HCl
28.05 g C2H4
1 mol HCl
36.46 g HCl
b.
c.
d.
e.
Since the mole ratio in the balanced equation is 1:1, the reactant with the
smaller number of moles is the limiting reactant: C2H4.
0.285 mol C2H4
0.285 mol C2H5Cl
10.6 g C2H5Cl
1 mol C2H5Cl
x
1 mol C2H4
x
64.51 g C2H5Cl
1 mol C2H5Cl
x
100%
=
=
0.285 mol C2H5Cl
=
18.4 g C2H5Cl
57.6 % percent yield
18.4 g C2H5Cl
5.88
a.
b.
c.
5.00 g CH4
x
15.0 g Cl2
x
1 mol CH4
16.05 g CH4
1 mol Cl2
70.90 g Cl2
=
0.312 mol CH4
=
0.212 mol Cl2
Since the mole ratio in the balanced equation is 1:2, (2)(0.312 mol) = 0.624
mol Cl2 would be needed to react with all of the CH4. There are only 0.212 mol
Cl2, however, so Cl2 is the limiting reactant.
0.212 mol Cl2
x
1 mol CH2Cl2
1 mol Cl2
=
0.212 mol CH2Cl2
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–26
d.
0.212mol CH2Cl2
15.6 g CH2Cl2
e.
x
84.93 g CH2Cl2
=
1 mol CH2Cl2
x
100%
=
18.0 g CH2Cl2
86.7 % percent yield
18.0 g CH2Cl2
5.89 A substance that is oxidized loses electrons, whereas an oxidizing agent gains electrons (it is
reduced).
5.90 A substance that is reduced gains electrons, whereas a reducing agent loses electrons (it is
oxidized).
5.91 The species that is oxidized loses one or more electrons. The species that is reduced gains one or
more electrons.
a.
Fe
oxidized
Cu2+
reduced
Fe
Fe2+
+
2 e–
Cu2+
b.
Cl2
reduced
2 I–
oxidized
2 I–
I2
+
2 e–
Cl2
+
2 e–
2 Cl–
c.
2 Na
oxidized
Cl2
reduced
2 Na
2 Na+
+
Cl2
+
2 e–
2 Cl–
2 e–
+
2 e–
Cu
5.92 The species that is oxidized loses one or more electrons. The species that is reduced gains one or
more electrons.
a.
Mg
oxidized
Fe2+
reduced
Mg
Mg2+
+
2 e–
Fe2+
b.
Cu2+
reduced
Sn
oxidized
Sn
Sn2+
+
2 e–
Cu2+
c.
4 Na
oxidized
O2
reduced
4 Na
4 Na+
+
O2
4 e–
2 e–
Fe
+
2 e–
Cu
+
4 e–
2 O2–
+
5.93 The oxidizing agent gains electrons (it is reduced). The reducing agent loses electrons (it is
oxidized).
Zn
+
Ag2O
ZnO
+
2 Ag
Zn
oxidized
reducing agent
Ag+
reduced
oxidizing agent
5.94 The oxidizing agent gains electrons (it is reduced). The reducing agent loses electrons (it is
oxidized).
Cd
+
Ni4+
Cd2+
+ Ni2+
Cd
oxidized
reducing agent
Ni4+
reduced
oxidizing agent
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–27
5.95 Acetylene is reduced because it gains hydrogen atoms.
5.96 Cl2 is reduced because it gains electrons.
5.97 Write the balanced equation and the half reactions.
2 MgO
2 Mg + O2
2 Mg2+ +
2 Mg
4 e–
O2
+
2 O2–
4 e–
5.98 Write the balanced equation and the half reactions.
4 Al +
4 Al
2 Al2O3
3 O2
4 Al3+ +
12 e–
3 O2
+
6 O2–
12 e–
5.99 Refer to prior solutions to answer each part.
C12H22O11(s)
+ H2O(l)
sucrose
C2H6O(l)
+ CO2(g)
ethanol
a. Calculate the molar mass as in Answer 5.59; the molar mass of sucrose = 342.3 g/mol.
b. Follow the steps in Example 5.2.
C12H22O11(s)
+
H2O(l)
4 C2H6O(l)
+
4 CO2(g)
c. 8 mol of ethanol are formed from 2 mol of sucrose.
d. 10 mol of water are needed to react with 10 mol of sucrose.
e. 101 g of ethanol are formed from 0.550 mol of sucrose.
f. 18.4 g of ethanol are formed from 34.2 g of sucrose.
g. 9.21 g ethanol
h. 13.6%
5.100 Refer to prior solutions to answer each part.
a. Calculate the molar mass as in Answer 5.59; the molar mass of diethyl ether = 74.1 g/mol.
b. Follow the steps in Example 5.2.
2 C2H6O(s)
ethanol
C4H10O(l) + H2O (l)
diethyl ether
c. 1 mol of diethyl ether is formed from 2 mol of ethanol.
d. 5 mol of water are formed from 10 mol of ethanol.
e. 20. g of diethyl ether are formed from 0.55 mol of ethanol.
f. 3.70 g of diethyl ether are formed from 4.60 g of ethanol.
g. 1.85 g diethyl ether
h. 97.3%
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chemical Reactions 5–28
5.101
a. 500 tablets
x
200. mg ibuprofen
1g
x
1000 mg
1 tablet
6.02 x 1023 molecules
b. 0.485 mol ibuprofen x
1 mol
=
x
1 mol ibuprofen
=
0.485 mol ibuprofen
206.3 g ibuprofen
2.92 x 1023 molecules
5.102
500. mg Mg(OH)2
1g
x
1000 mg
500. mg Al(OH)3
1g
x
1000 mg
1 mol Mg(OH)2
x
8.57 x 10-3 mol Mg(OH)2
=
58.33 g Mg(OH)2
1 mol Al(OH)3
x
= 6.41x 10-3 mol Al(OH)3
78.01 g Al(OH)3
5.103
a. 20 cig x 1.93 mg
1 cig
1g
x
1 mol nicotine
x
1000 mg
6.02 x 1023 molecules
b. 2.38 x 10–4 mol nicotine x
=
2.38 x 10–4 mol nicotine
162.3 g nicotine
1 mol
=
1.43 x 1020 molecules
5.104
5 lb
454 g
x
1 lb
1 mol
x
=
7 mol sucrose
342.3 g
5.105
2400 mg
x
1g
1 mol
x
1000 mg
x
22.99 g
6.02 x 1023 ions
1 mol
= 6.3 x 1022 ions
5.106
250 g
14 mol
x
x
1 mol
=
14 mol water
18.0 g
6.022 x 1023 molecules
=
8.4 x 1024 molecules water
1 mol
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 5–29
5.107
H
2
Cl
H
C
C
C
C
C
H
C
C
H
H
H
+
C2HCl3O
Cl
C
C
C
H
H
chlorobenzene
C
C
H
CCl3 C
C
C
C
C
H
H
H
C
H
Cl
+
H2O
C
H
DDT
C14H9Cl5
112.6 g/mol
a. Calculate the molar mass as in Answer 5.59; the molar mass of DDT = 354.5 g/mol.
b. 18 g of DDT would be formed from 0.10 mol of chlorobenzene.
c. 17.8 g is the theoretical yield of DDT in grams from 11.3 g of chlorobenzene.
d. 84.3%
5.108
Refer to prior solutions to answer each part.
a. Calculate the molar mass as in Answer 5.59; the molar mass of linolenic acid = 278.5 g/mol.
C18H36O2
b. C18H30O2 + 3 H2
2
C
H
O
36 CO2 + 30 H2O
c.
49 O2
18 30 2 +
d. 10.2 grams of C18H36 O2 will be formed.
5.109
Use conversion factors to answer the questions about dioxin.
a. 70. kg x
b.
5.110
3.0 x 10–2 mg
1 kg
2.1 x 10–3 g dioxin x
x
1g
1000 mg
1 mol
322.0 g
x
= 2.1 x 10–3 g dioxin
6.02 x 1023 molecules
1 mol
=
3.9 x 1018 molecules
Pb is the reducing agent. It is oxidized from Pb to Pb2+. PbO2 is the oxidizing agent. Pb is
reduced from +4 in PbO2 to +2 in PbSO4.
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.