Download Chapter 3 Solutions - Bremerton School District

CHAPTER THREE
STOICHIOMETRY
Questions
18.
The two major isotopes of boron are 10 B and 11 B. The listed mass of 10.81 is the average mass of a
very large number of boron atoms.
19.
The molecular formula tells us the actual number of atoms of each element in a molecule (or formula
unit) of a compound. The empirical formula tells only the simplest whole number ratio of atoms of
each element in a molecule. The molecular formula is a whole number multiple of the empirical
formula. If that multiplier is one, the molecular and empirical formulas are the same. For example,
both the molecular and empirical formulas of water are H2 O. They are the same. For hydrogen
peroxide, the empirical formula is OH; the molecular formula is H2 O2 .
20.
Side reactions may occur. For example, in the combustion of CH4 (methane) to CO2 and H2O, some
CO is also formed. Also, some reactions only go part way to completion and reach a state of
equilibrium where both reactants and products are present (see Ch. 13).
Exercises
Atomic Masses and the Mass Spectrometer
21.
A = atomic mass = 0.7899(23.9850 amu) + 0.1000(24.9858 amu) + 0.1101(25.9826 amu)
A = 18.95 amu + 2.499 amu + 2.861 amu = 24.31 amu
22.
A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766)
A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 amu; From the periodic table, the element is Pb.
23.
Let x = % of 151 Eu and y = % of 153 Eu, then x + y = 100 and y = 100 - x.
151.96 =
15196 = 150.9196 x + 15292.09 - 152.9209 x, -96 = -2.0013 x
x = 48%; 48% 151 Eu and 100 - 48 = 52% 153 Eu
29
30
24.
CHAPTER 3
STOICHIOMETRY
Let A = mass of 185 Re: 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 - 117.0 = 0.3740(A)
A=
= 185 amu (A = 184.95 amu without rounding to proper significant figures.)
25.
There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two
isotopes, differing in mass by two mass units. The peak at 157.84 corresponds to a Br2 molecule
composed of two atoms of the lighter isotope. This isotope has mass equal to 157.84/2 or 78.92. This
corresponds to 79 Br. The second isotope is 81 Br with mass equal to 161.84/2 = 80.92. The peaks in
the mass spectrum correspond to 79 Br2 , 79 Br81 Br, and 81 Br2 in order of increasing mass. The intensities
of the highest and lowest mass tell us the two isotopes are present in about equal abundance. The
actual abundance is 50.69% 79 Br and 49.31% 81 Br. The calculation of the abundance from the mass
spectrum is beyond the scope of this text.
26.
GaAs can be either 69 GaAs or 71 GaAs. The mass spectrum for GaAs will have 2 peaks at 144
(= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60:40 or 3:2.
Ga2 As2 can be 69 Ga2 As2 , 69 Ga71 GaAs2 , or 71 Ga2 As2 . The mass spectrum will have 3 peaks at 288, 290,
and 292 with intensities in the ratio of 36:48:16 or 9:12:4. We get this ratio from the following
probability table:
69
Ga (0.60)
71
Ga (0.40)
69
Ga (0.60)
0.36
0.24
71
Ga (0.40)
0.24
0.16
CHAPTER 3
STOICHIOMETRY
31
Moles and Molar Masses
27.
When more than one conversion factor is necessary to determine the answer, we will usually put all
the conversion factors into one calculation instead of determining intermediate answers. This method
reduces round-off error and is a time saver.
= 4.64 × 10-20 g Fe
500. atoms Fe ×
28.
500.0 g Fe ×
8.953 mol Fe ×
29.
1.00 carat ×
30.
5.0 × 1021 atoms C ×
8.3 × 10-3 mol C ×
31.
= 8.953 mol Fe
= 5.391 × 1024 atoms Fe
= 1.00 × 1022 atoms C
= 8.3 × 10-3 mol C
= 0.10 g C
Al2 O3 : 2(26.98) + 3(16.00) = 101.96 g/mol
Na3 AlF6 : 3(22.99) + 1(26.98) + 6(19.00) = 209.95 g/mol
32.
HFC - 134a, CH2 FCF3 : 2(12.01) + 2(1.008) + 4(19.00) = 102.04 g/mol
HCFC-124, CHClFCF3 : 2(12.01) + 1(1.008) + 1(35.45) + 4(19.00) = 136.48 g/mol
33.
a. The formula is NH3 . 14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol
b. The formula is N2 H4 . 2(14.01) + 4(1.008) = 32.05 g/mol
c. (NH4 )2Cr2O7: 2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 g/mol
34.
a. The formula is P4 O6 . 4(30.97 g/mol) + 6(16.00 g/mol) = 219.88 g/mol
b. Ca3 (PO4 )2 : 3(40.08) + 2(30.97) + 8(16.00) = 310.18 g/mol
c. Na2 HPO4: 2(22.99) + 1(1.008) + 1(30.97) + 4(16.00) = 141.96 g/mol
35.
a. 1.00 g NH3 ×
= 0.0587 mol NH3
32
CHAPTER 3
b. 1.00 g N2 H4 ×
= 0.0312 mol N2H4
= 3.97 × 10-3 mol (NH4 )2 Cr2 O7
c. 1.00 g (NH4 )2 Cr2 O7 ×
36.
a. 1.00 g P4 O6 ×
b. 1.00 g Ca3 (PO4 )2 ×
c. 1.00 g Na2 HPO4 ×
37.
a. 5.00 mol NH3 ×
b. 5.00 mol N2 H4 ×
= 4.55 × 10-3 mol P4 O6
= 3.22 × 10-3 mol Ca3 (PO4 )2
= 7.04 × 10-3 mol Na2 HPO4
= 85.2 g NH3
= 160. g N2H4
c. 5.00 mol (NH4 )2 Cr2 O7 ×
38.
a. 5.00 mol P4 O6 ×
b. 5.00 mol Ca3 (PO4 )2 ×
c. 5.00 mol Na2 HPO4 ×
39.
= 1260 g (NH4 )2 Cr2 O7
= 1.10 × 103 g P4 O6
= 1.55 × 103 g Ca3(PO4)2
= 7.10 × 102 g Na2 HPO4
Chemical formulas give atom ratios as well as mol ratios.
a. 5.00 mol NH3 ×
b. 5.00 mol N2 H4 ×
= 70.1 g N
= 140. g N
c. 5.00 mol (NH4 )2 Cr2 O7 ×
40.
a. 5.00 mol P4 O6 ×
STOICHIOMETRY
= 140. g N
= 619 g P
CHAPTER 3
STOICHIOMETRY
b. 5.00 mol Ca3 (PO4 )2 ×
c. 5.00 mol Na2 HPO4 ×
41.
a. 1.00 g NH3 ×
b. 1.00 g N2 H4 ×
33
= 310. g P
= 155 g P
= 3.54 × 1022 molecules NH3
= 1.88 × 1022 molecules N2 H4
c. 1.00 g (NH4 )2 Cr2 O7 ×
= 2.39 × 1021 formula units (NH4 )2 Cr2 O7
42.
a. 1.00 g P4 O6 ×
= 2.74 × 1021 molecules P4 O6
b. 1.00 g Ca3 (PO4 )2 ×
= 1.94 × 1021 formula units Ca3 (PO4 )2
c. 1.00 g Na2 HPO4 ×
= 4.24 × 1021 formula units Na2 HPO4
43.
Using answers from Exercise 41:
a. 3.54 × 1022 molecules NH3 ×
b. 1.88 × 1022 molecules N2 H4 ×
= 3.54 × 1022 atoms N
= 3.76 × 1022 atoms N
c. 2.39 × 1021 formula units (NH4 )2 Cr2 O7 ×
44.
= 4.78 × 1021 atoms N
Using answers from Exercise 42:
a. 2.74 × 1021 molecules P4 O6 ×
= 1.10 × 1022 atoms P
34
CHAPTER 3
b. 1.94 × 1021 formula units Ca3 (PO4 )2 ×
= 3.88 × 1021 atoms P
c. 4.24 × 1021 formula units Na2 HPO4 ×
45.
= 2.839 × 10-3 mol
2.839 × 10-3 mol ×
= 1.710 × 1021 molecules
a. 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol
= 2.78 × 10-3 mol
b. 500. mg ×
2.78 × 10-3 mol ×
47.
= 4.24 × 1021 atoms P
Molar mass of C6 H8 O6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol
500.0 mg ×
46.
= 1.67 × 1021 molecules
a. 2.49 × 1020 molecules CO ×
= 4.13 × 10-4 mol CO
= 9.40 ×10-2 mol CuSO4
b. 15.0 g CuSO4 ×
= 1.661 × 10-22 mol H2 SO4
c. 100 molecules H2 SO4 ×
= 6.592 × 10-5 mol K2O
d. 6.210 mg K2 O ×
48.
a. 150.0 g Fe2 O3 ×
= 0.9393 mol Fe2 O3
b. 10.0 mg NO2 ×
49.
STOICHIOMETRY
c. 1.5 × 1016 molecules BF3 ×
a. 1.27 mmol CO2 ×
b. 2.00 × 1022 molecules NCl3 ×
= 2.17 × 10-4 mol NO2
= 2.5 × 10-8 mol BF3
= 5.59 × 10-2 g CO2
= 4.00 g NCl3
CHAPTER 3
STOICHIOMETRY
35
c. 0.00451 mol (NH4 )2 CO3 ×
= 0.433 g (NH4 )2 CO3
= 4.653 × 10-23 g N2
d. 1 molecule N2 ×
= 1.00 × 104 g CuSO4
e. 62.7 mol CuSO4 ×
50.
a. A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to show
how these conversion factors can be used.
Molar mass of C2 H5 O2 N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol
5.00 g C2 H5 O2 N ×
×
b. Molar mass of Mg3 N2 = 3(24.31) + 2(14.01) = 100.95 g/mol
5.00 g Mg3 N2 ×
= 5.97 × 1022 atoms N
c. Molar mass of Ca(NO3 )2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol
5.00 g Ca(NO3 )2 ×
= 3.67 × 1022 atoms N
d. Molar mass of N2 O4 = 2(14.01) + 4(16.00) = 92.02 g/mol
= 6.54 × 1022 atoms N
5.00 g N2 O4 ×
51.
a. 14 mol C
+ 18 mol H
+ 2 mol N
+ 5 mol O
= 294.30 g/mol
b. 10.0 g aspartame ×
= 3.40 × 10-2 mol
36
CHAPTER 3
c. 1.56 mol ×
STOICHIOMETRY
= 459 g
= 1.0 × 1019 molecules
d. 5.0 mg ×
e. The chemical formula tells us that 1 molecule of aspartame contains two atoms of N. The
chemical formula also says that 1 mol of aspartame contains two mol of N.
1.2 g aspartame ×
= 4.9 × 1021 atoms of nitrogen
f.
1.0 × 109 molecules ×
= 4.9 × 10-13 g or 490 fg
= 4.887 × 10-22 g
g. 1 molecule aspartame ×
52.
a. 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol
b. 500.0 g ×
= 3.023 mol
c. 2.0 × 10-2 mol ×
= 3.3 g
d. 5.0 g C2 H3 Cl3 O2 ×
= 5.5 × 1022 atoms of chlorine
e. 1.0 g Cl ×
f.
= 1.6 g chloral hydrate
= 1.373 × 10-19 g
500 molecules ×
Percent Composition
53.
In 1 mole of YBa2 Cu3 O7 , there are 1 mole of Y, 2 moles of Ba, 3 moles of Cu and 7 moles of O.
Molar mass = 1 mol Y
+ 2 mol Ba
+ 3 mol Cu
+ 7 mol O
CHAPTER 3
STOICHIOMETRY
37
Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol
%Y =
× 100 = 13.35% Y; %Ba =
%Cu =
54.
× 100 = 41.22% Ba
× 100 = 28.62% Cu; %O =
× 100 = 16.81% O
a. C3 H4 O2 : Molar mass = 3(12.01) + 4(1.008) + 2(16.00) = 36.03 + 4.032 + 32.00 = 72.06 g/mol
%C =
× 100 = 50.00% C; %H =
× 100 = 5.595% H
%O = 100.00 - (50.00 + 5.595) = 44.41% O or %O =
× 100 = 44.41% O
b. C4 H6 O2 : Molar mass = 4(12.01) + 6(1.008) + 2(16.00) = 48.04 + 6.048 + 32.00 = 86.09 g/mol
%C =
× 100 = 55.80% C; %H =
× 100 = 7.025% H
%O = 100.00 - (55.80 + 7.025) = 37.18% O
c. C3 H3 N: Molar mass = 3(12.01) + 3(1.008) + 1(14.01) = 36.03 + 3.024 + 14.01 = 53.06 g/mol
55.
%C =
× 100 = 67.90% C;
%N =
× 100 = 26.40% N or %N = 100.00 - (67.90 + 5.699) = 26.40% N
a. NO: %N =
b. NO2 : %N =
c. N2 O4 : %N =
d. N2 O: %N =
%H =
× 100 = 5.699% H
× 100 = 46.68% N
× 100 = 30.45% N
× 100 = 30.45% N
× 100 = 63.65% N
The order from lowest to highest mass percentage of nitrogen is: NO2 = N2 O4 < NO < N2 O.
56.
C8 H10 N4 O2 : molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol
%C =
× 100 =
× 100 = 49.47% C
38
CHAPTER 3
STOICHIOMETRY
C12 H22 O11 : molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol
%C =
× 100 = 42.10% C
C2 H5 OH: molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol
%C =
× 100 = 52.14% C
The order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11) < caffeine
(C8 H10 N4 O2 ) < ethanol (C2 H5 OH)
57.
There are many valid methods to solve this problem. We will assume 100.00 g of compound; then
determine from the information in the problem how many mol of compound equals 100.00 g of
compound. From this information, we can determine the mass of one mol of compound (the molar
mass) by setting up a ratio. Assuming 100.00 g cyanocobalamin:
mol cyanocobalamin = 4.34 g Co ×
= 7.36 × 10-2 mol cyanocobalamin
, x = molar mass = 1360 g/mol
58.
There are 0.390 g Cu for every 100.00 g of fungal laccase. Assuming 100.00 g fungal laccase:
= 1.53 × 10-3 mol
mol fungal laccase = 0.390 g Cu ×
, x = molar mass = 6.54 × 104 g/mol
Empirical and Molecular Formulas
59.
a. Molar mass of CH2 O = 1 mol C
+ 2 mol H
+ 1 mol O
%C =
× 100 = 39.99% C; %H =
= 30.03 g/mol
× 100 = 6.713% H
CHAPTER 3
STOICHIOMETRY
%O =
39
× 100 = 53.28% O
or %O = 100.00 - (39.99 + 6.713) = 53.30%
b. Molar Mass of C6 H12 O6 = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol
%C =
× 100 = 40.00%; %H =
× 100 = 6.714%
%O = 100.00 - (40.00 + 6.714) = 53.29%
c. Molar mass of HC2 H3 O2 = 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol
%C =
× 100 = 40.00%;
%H =
× 100 = 6.714%
%O = 100.00 - (40.00 + 6.714) = 53.29%
60.
All three compounds have the same empirical formula, CH2 O, and different molecular formulas. The
composition of all three in mass percent is also the same (within rounding differences). Therefore,
elemental analysis will give us only the empirical formula.
61.
a. The molecular formula is N2O4. The smallest whole number ratio of the atoms (the empirical
formula) is NO2 .
b. Molecular formula: C3H6; empirical formula = CH2
c. Molecular formula: P4O10; empirical formula = P2O5
d. Molecular formula: C6 H12 O6 ; empirical formula = CH2 O
62.
a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g
= 4.000; So the molecular formula is (SNH)4 or S4 N4 H4 .
b. NPCl2 : Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol
= 3.0000; Molecular formula is (NPCl2 )3 or N3 P3 Cl6 .
c. CoC4 O4 : 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol
= 2.0000; Molecular formula: Co2 C8 O8
d. SN: 32.07 + 14.01 = 46.08 g/mol;
= 4.000; Molecular formula: S4N4
40
63.
CHAPTER 3
STOICHIOMETRY
Out of 100.0 g of the pigment, there are:
59.9 g Ti ×
= 1.25 mol Ti; 40.1 g O ×
= 2.51 mol O
Empirical formula = TiO2 since mol O to mol Ti are in a 2:1 mol ratio (2.51/1.25 = 2.01).
64.
Out of 100.00 g of adrenaline, there are:
56.79 g C ×
= 4.729 mol C; 6.56 g H ×
28.37 g O ×
= 6.51 mol H
= 1.773 mol O; 8.28 g N ×
= 0.591 mol N
Dividing each mol value by the smallest number:
= 8.00;
= 11.0;
= 3.00;
= 1.00
This gives adrenaline an empirical formula of C8H11O3N.
65.
Compound I: mass O = 0.6498 g Hgx Oy - 0.6018 g Hg = 0.0480 g O
= 3.000 × 10-3 mol Hg
0.6018 g Hg ×
0.0480 g O ×
= 3.00 × 10-3 mol O
The mol ratio between Hg and O is 1:1, so the empirical formula of compound I is HgO.
Compound II: mass Hg = 0.4172 g Hgx Oy - 0.016 g O = 0.401 g Hg
= 2.00 × 10-3 mol Hg; 0.016 g O ×
0.401 g Hg ×
= 1.0 × 10-3 mol O
The mol ratio between Hg and O is 2:1, so the empirical formula is Hg2 O.
66.
1.121 g N ×
= 8.001 × 10-2 mol N; 0.161 g H ×
0.480 g C ×
= 4.00 × 10-2 mol C; 0.640 g O ×
= 1.60 × 10-1 mol H
= 4.00 × 10-2 mol O
Dividing all mol values by the smallest number:
= 2.00;
= 4.00;
= 1.00
Empirical formula = N2 H4 CO
67.
Out of 100.0 g compound: 30.4 g N ×
= 2.17 mol N
CHAPTER 3
STOICHIOMETRY
41
%O = 100.0 - 30.4 = 69.6% O; 69.6 g O ×
= 1.00;
= 4.35 mol O
= 2.00; Empirical formula is NO2 .
The empirical formula mass of NO2 . 14 + 2(16) = 46 g/mol.
= 2.0; Therefore, the molecular formula is N2 O4 .
68.
Out of 100.0 g, there are:
69.6 g S ×
= 2.17 mol S; 30.4 g N ×
= 2.17 mol N
Empirical formula is SN since mol values are in a 1:1 mol ratio.
The empirical formula mass of SN is ~ 46 g. Since
69.
= 4.0, the molecular formula is S4N4.
Assuming 100.00 g of compound (mass hydrogen = 100.00 g - 49.31 g C - 43.79 g O = 6.90 g H):
49.31 g C ×
= 4.106 mol C; 6.90 g H ×
43.79 g O ×
= 2.737 mol O
= 6.85 mol H
Dividing all mole values by 2.737 gives:
= 1.500;
= 2.50;
= 1.000
Since a whole number ratio is required, the empirical formula is C3H5O2.
The empirical formula mass is: 3(12.01) + 5(1.008) +2(16.00) = 73.07 g/mol
= 1.999; molecular formula = (C3 H5 O2 )2 = C6 H10 O4
70.
Assuming 100.00 g of compound (mass oxygen = 100.00 g - 41.39 g C - 3.47 g H = 55.14 g O):
41.39 g C ×
= 3.446 mol C; 3.47 g H ×
55.14 g O ×
= 3.446 mol O
= 3.44 mol H
All are the same mol values so the empirical formula is CHO. The empirical formula mass is
12.01 + 1.008 + 16.00 = 29.02 g/mol.
42
CHAPTER 3
molar mass =
STOICHIOMETRY
= 116 g/mol
= 4.00; molecular formula = (CHO)4 = C4 H4 O4
71.
When combustion data are given, it is assumed that all the carbon in the compound ends up as carbon
in CO2 and all the hydrogen in the compound ends up as hydrogen in H2O. In the sample of propane
combusted, the moles of C and H are:
mol C = 2.641 g CO2 ×
= 0.06001 mol C
mol H = 1.442 g H2 O ×
= 0.1600 mol H
= 2.666
Multiplying this ratio by three gives the empirical formula of C3H8.
72.
This compound contains nitrogen, and one way to determine the amount of nitrogen in the compound
is to calculate composition by mass percent. We assume that all of the carbon in 33.5 mg CO2 came
from the 35.0 mg of compound and all of the hydrogen in 41.1 mg H2 O came from the 35.0 mg of
compound.
3.35 × 10-2 g CO2 ×
%C =
= 9.14 × 10-3 g C
× 100 = 26.1% C
4.11 × 10-2 g H2 O ×
%H =
= 4.60 × 10-3 g H
× 100 = 13.1% H
The mass percent of nitrogen is obtained by difference:
%N = 100.0 - (26.1 + 13.1) = 60.8% N
Now perform the empirical formula determination by first assuming 100.0 g of compound. Out of
100.0 g of compound, there are:
26.1 g C ×
= 2.17 mol C; 13.1 g H ×
= 13.0 mol H
CHAPTER 3
STOICHIOMETRY
60.8 g N ×
43
= 4.34 mol N
Dividing all mol values by 2.17 gives:
= 5.99;
= 2.00
The empirical formula is CH6 N2 .
73.
The combustion data allow determination of the amount of hydrogen in cumene. One way to determine
the amount of carbon in cumene is to determine the mass percent of hydrogen in the compound from
the data in the problem; then determine the mass percent of carbon by difference (100.0 - mass %H
= mass %C).
42.8 mg H2 O ×
= 4.79 mg H
%H =
× 100 = 10.1% H; %C = 100.0 - 10.1 = 89.9% C
Now solve this empirical formula problem. Out of 100.0 g cumene, we have:
89.9 g C ×
= 1.34 .
= 7.49 mol C; 10.1 g H ×
= 10.0 mol H
, i.e., mol H to mol C are in a 4:3 ratio. Empirical formula = C3 H4
Empirical formula mass . 3(12) + 4(1) = 40 g/mol
The molecular formula is (C3 H4 )3 or C9 H12 since the molar mass will be between 115 and 125 g/mol
(molar mass . 3 × 40 g/mol = 120 g/mol).
74.
First, we will determine composition by mass percent:
16.01 mg CO2 ×
%C =
= 4.369 mg C
× 100 = 40.91% C
4.37 mg H2 O ×
%H =
= 0.489 mg H
× 100 = 4.58% H; %O = 100.00 - (40.91 + 4.58) = 54.51% O
So, in 100.00 g of the compound, we have:
40.91 g C ×
= 3.406 mol C; 4.58 g H ×
= 4.54 mol H
44
CHAPTER 3
54.51 g O ×
STOICHIOMETRY
= 3.407 mol O
Dividing by the smallest number:
= 1.33 =
; the empirical formula is C3 H4 O3 .
The empirical formula mass of C3H4O3 is . 3(12) + 4(1) + 3(16) = 88 g.
Since
= 2.0, then the molecular formula is C6 H8 O6 .
Balancing Chemical Equations
75.
When balancing reactions, start with elements that appear in only one of the reactants and one of
the products, then go on to balance the remaining elements.
a. Fe + O2 ÷ Fe2O3. Balancing Fe first, then O, gives: 2 Fe + 3/2 O2 ÷ Fe2O3. The best
balanced equation contains the smallest whole numbers. To convert to whole numbers, multiply
each coefficient by two, which gives: 4 Fe(s) + 3 O2 (g) ÷ 2 Fe2O3(s)
b. Ca + H2 O ÷ Ca(OH)2 + H2 ; Calcium is already balanced, so concentrate on oxygen next.
Balancing O gives: Ca(s) + 2 H2 O(l) ÷ Ca(OH)2(aq) + H2 (g). The equation is balanced.
Note: Hydrogen is the most difficult element to balance since it appears in both products. It is
generally easiest to save these atoms for last when balancing an equation.
c. Ba(OH)2 + H2 SO4 ÷ BaSO4 + H2O; Ba and S are already balanced. There are 6 O atoms
on the reactant side and, in order to get 6 O atoms on the product side, we will need 2 H2O
molecules. The balanced equation is: Ba(OH)2 (aq) + H2SO4(aq)÷ BaSO4(s) + 2 H2O(l).
76.
a. C6 H12 O6 (s) + O2 (g) ÷ CO2 (g) + H2 O(g)
Balance C atoms: C6 H12 O6 + O2 ÷ 6 CO2 + H2 O
Balance H atoms: C6 H12 O6 + O2 ÷ 6 CO2 + 6 H2 O
Lastly, balance O atoms: C6 H12 O6 (s) + 6 O2 (g) ÷ 6 CO2 (g) + 6 H2 O(g)
The equation is balanced.
b. Fe2 S3 (s) + HCl(g) ÷ FeCl3 (s) + H2 S(g)
Balance Fe atoms: Fe2 S3 + HCl ÷ 2 FeCl3 + H2S
Balance S atoms: Fe2 S3 + HCl ÷ 2 FeCl3 + 3 H2 S
There are 6 H and 6 Cl on right, so balance with 6 HCl on left:
CHAPTER 3
STOICHIOMETRY
45
Fe2S3(s) + 6 HCl(g) ÷ 2 FeCl3(s) + 3 H2S(g). Equation is balanced.
c. CS2 (l) + NH3 (g) ÷ H2 S(g) + NH4 SCN(s)
C and S balanced; balance N:
CS2 + 2 NH3 ÷ H2 S + NH4 SCN
H is also balanced. So: CS2 (l) + 2 NH3 (g) ÷ H2 S(g) + NH4 SCN(s)
77.
a. Cu(s) + 2 AgNO3 (aq) ÷ 2 Ag(s) + Cu(NO3 )2 (aq)
b. Zn(s) + 2 HCl(aq) ÷ ZnCl2 (aq) + H2 (g)
c. Au2 S3 (s) + 3 H2 (g) ÷ 2 Au(s) + 3 H2 S(g)
78.
a. 3 Ca(OH)2 (aq) + 2 H3 PO4 (aq) ÷ 6 H2 O(l) + Ca3 (PO4 )2 (s)
b. Al(OH)3 (s) + 3 HCl(aq) ÷ AlCl3 (aq) + 3 H2 O(l)
c. 2 AgNO3 (aq) + H2 SO4 (aq) ÷ Ag2 SO4 (s) + 2 HNO3 (aq)
79.
a. The formulas of the reactants and products are C6 H6 (l) + O2 (g) ÷ CO2 (g) + H2 O(g).
To balance this combustion reaction, notice that all of the carbon in C6 H6 has to end up as carbon
in CO2 and all of the hydrogen in C6 H6 has to end up as hydrogen in H2 O. To balance C and H,
we need 6 CO2 molecules and 3 H2 O molecules for every 1 molecule of C6 H6 . We do oxygen last.
Since we have 15 oxygen atoms in 6 CO2 molecules and 3 H2O molecules, we need 15/2 O2
molecules in order to have 15 oxygen atoms on the reactant side.
C6 H6 (l) +
O2 (g) ÷ 6 CO2 (g) + 3 H2 O(g); Multiply by two to give whole numbers.
2 C6 H6 (l) + 15 O2 (g) ÷ 12 CO2 (g) + 6 H2 O(g)
b. The formulas of the reactants and products are C4 H10 (g) + O2 (g) ÷ CO2 (g) + H2 O(g).
C4 H10 (g) +
O2 (g) ÷ 4 CO2 (g) + 5 H2 O(g); Multiply by two to give whole numbers.
2 C4 H10 (g) + 13 O2 (g) ÷ 8 CO2 (g) + 10 H2 O(g)
c. C12 H22 O11 (s) + 12 O2 (g) ÷ 12 CO2 (g) + 11 H2 O(g)
d. 2 Fe(s) +
e. 2 FeO(s) +
O2 (g) ÷ Fe2 O3 (s); For whole numbers: 4 Fe(s) + 3 O2 (g) ÷ 2 Fe2 O3 (s)
O2 (g) ÷ Fe2 O3 (s); For whole numbers, multiply by two.
4 FeO(s) + O2 (g) ÷ 2 Fe2 O3 (s)
80.
a. 16 Cr(s) + 3 S8 (s) ÷ 8 Cr2 S3 (s)
46
CHAPTER 3
STOICHIOMETRY
b. 2 NaHCO3 (s) ÷ Na2 CO3 (s) + CO2 (g) + H2 O(g)
c. 2 KClO3 (s) ÷ 2 KCl(s) + 3 O2 (g)
d. 2 Eu(s) + 6 HF(g) ÷ 2 EuF3 (s) + 3 H2 (g)
81.
a. SiO2 (s) + C(s) ÷ Si(s) + CO(g)
Balance oxygen atoms: SiO2 + C ÷ Si + 2 CO
Balance carbon atoms: SiO2 (s) + 2 C(s) ÷ Si(s) + 2 CO(g)
b. SiCl4 (l) + Mg(s) ÷ Si(s) + MgCl2(s)
Balance Cl atoms: SiCl4 + Mg ÷ Si + 2 MgCl2
Balance Mg atoms: SiCl4 (l) + 2 Mg(s) ÷ Si(s) + 2 MgCl2 (s)
c. Na2 SiF6 (s) + Na(s) ÷ Si(s) + NaF(s)
Balance F atoms:
Na2 SiF6 + Na ÷ Si + 6 NaF
Balance Na atoms: Na2 SiF6 (s) + 4 Na(s) ÷ Si(s) + 6 NaF(s)
82.
Unbalanced equation:
CaF2 C3Ca3 (PO4 )2 (s) + H2 SO4 (aq) ÷ H3 PO4 (aq) + HF(aq) + CaSO4 C2H2 O(s)
Balancing Ca2+ , F-, and PO4 3-:
CaF2 C3Ca3 (PO4 )2 (s) + H2 SO4 (aq) ÷ 6 H3 PO4 (aq) + 2 HF(aq) + 10 CaSO4 C2H2 O(s)
On the right-hand side there are 20 extra hydrogen atoms, 10 extra sulfates, and 20 extra water
molecules. We can balance the hydrogen and sulfate with 10 sulfuric acid molecules. The extra
waters came from the water in the sulfuric acid solution. The balanced equation is:
CaF2 C3Ca3 (PO4 )2 (s) + 10 H2 SO4 (aq) + 20 H2 O(l) ÷ 6 H3 PO4 (aq) + 2 HF(aq) + 10 CaSO4C2H2O(s)
83.
C12 H22 O11 (aq) + H2 O(l) ÷ 4 C2 H5 OH(aq) + 4 CO2 (g)
84.
CaSiO3 (s) + 6 HF(aq) ÷ CaF2 (aq) + SiF4 (g) + 3 H2 O(l)
Reaction Stoichiometry
85.
The stepwise method to solve stoichiometry problems is outlined in the text. Instead of calculating
intermediate answers for each step, we will combine conversion factors into one calculation. This
CHAPTER 3
STOICHIOMETRY
47
practice reduces round-off error and saves time.
The balanced reaction is: (NH4 )2 Cr2 O7 (s) ÷ Cr2O3(s) + N2(g) + 4 H2O(g)
= 4.28 × 10-2 mol (NH4 )2 Cr2 O7
10.8 g (NH4 )2 Cr2 O7 ×
4.28 × 10-2 mol (NH4 )2 Cr2 O7 ×
= 6.51 g Cr2 O3
4.28 × 10-2 mol (NH4 )2 Cr2 O7 ×
= 1.20 g N2
4.28 × 10-2 mol (NH4 )2 Cr2 O7 ×
86.
= 3.09 g H2 O
Fe2 O3 (s) + 2 Al(s) ÷ 2 Fe(l) + Al2 O3 (s)
15.0 g Fe ×
= 0.269 mol Fe;
0.269 mol Fe ×
0.269 mol Fe ×
= 21.5 g Fe2 O3
0.269 mol Fe ×
= 13.7 g Al2 O3
= 7.26 g Al
87.
1.000 kg Al ×
88.
a. Ba(OH)2 C8H2 O(s) + 2 NH4 SCN(s) ÷ Ba(SCN)2(s) + 10 H2O(l) + 2 NH3(g)
b. 6.5 g Ba(OH)2 C8H2 O ×
0.021 mol Ba(OH)2 C8H2 O ×
89.
= 4355 g
= 0.0206 mol = 0.021 mol
= 3.2 g NH4 SCN
1.0 ton CuO ×
= 7.2 × 104 g coke
90.
1.0 × 104 kg waste ×
48
CHAPTER 3
STOICHIOMETRY
= 3.4 × 104 g tissue if all NH4 + converted
Since only 95% of the NH4 + ions react:
mass of tissue = (0.95) (3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue
91.
a. Molar mass = 195.1 + 2(14.01) + 6(1.008) + 2(35.45) = 300.1 g/mol
% Pt =
× 100 = 65.01% Pt; % N =
× 100 = 9.337% N
%H=
× 100 = 2.015% H; % Cl =
× 100 = 23.63% Cl
65.01% Pt; 9.337% N; 2.015% H; 23.63% Cl
b. 100. g K2 PtCl4 ×
= 72.3 g Pt(NH3 )2 Cl2
100. g K2 PtCl4 ×
92.
= 35.9 g KCl
a. 1.00 × 102 g C7 H6 O3 ×
= 73.9 g C4 H6 O3
b. 1.00 × 102 g C7 H6 O3 ×
= 1.30 × 102 g aspirin
Limiting Reactants and Percent Yield
93.
a. Mg(s) + I2 (s) ÷ MgI2 (s)
From the balanced equation, 100 molecules of I2 reacts completely with 100 atoms of Mg. We
have a stoichiometric mixture. Neither is limiting.
b. 150 atoms Mg ×
= 150 molecules I2 needed
We need 150 molecules I2 to react completely with 150 atoms Mg; we only have 100 molecules
I2 . Therefore, I2 is limiting.
c. 200 atoms Mg ×
d. 0.16 mol Mg ×
= 200 molecules I2.; Mg is limiting since 300 molecules I2
are present.
= 0.16 mol I2 ; Mg is limiting since 0.25 mol I2 are present.
CHAPTER 3
STOICHIOMETRY
49
e. 0.14 mol Mg ×
= 0.14 mol I2 needed; Stoichiometric mixture. Neither is limiting.
f.
= 0.12 mol I2 needed; I2 is limiting since only 0.08 mol I2 are
present.
0.12 mol Mg ×
g. 6.078 g Mg ×
= 63.46 g I2
Stoichiometric mixture. Neither is limiting.
h. 1.00 g Mg ×
= 10.4 g I2
10.4 g I2 needed, but we only have 2.00 g. I2 is limiting.
i.
94.
From h above, we calculated that 10.4 g I2 will react completely with 1.00 g Mg. We have 20.00
g I2 . I2 is in excess. Mg is limiting.
2 H2 (g) + O2 (g) ÷ 2 H2 O(g)
a. 50 molecules H2 ×
= 25 molecules O2
Stoichiometric mixture. Neither is limiting.
b. 100 molecules H2 ×
= 50 molecules O2 ; O2 is limiting since only 40 molecules
O2 are present.
c. From b, 50 molecules of O2 will react completely with 100 molecules of H2. We have 100
molecules (an excess) of O2 . So, H2 is limiting.
d. 0.50 mol H2 ×
= 0.25 mol O2 ; H2 is limiting since 0.75 mol O2 are present.
e. 0.80 mol H2 ×
= 0.40 mol O2 ; H2 is limiting since 0.75 mol O2 are present.
f.
1.0 g H2 ×
= 0.25 mol O2
Stoichiometric mixture, neither is limiting.
g. 5.00 g H2 ×
95.
a. 10.0 g Hg ×
= 39.7 g O2 ; H2 is limiting since
56.00 g O2 are present.
= 4.99 × 10-2 mol Hg
50
CHAPTER 3
STOICHIOMETRY
= 5.63 × 10-2 mol Br2
9.00 g Br2 ×
The required mol ratio from the balanced equation is 1 mol Br2 to 1 mol Hg. The actual mol ratio
is:
= 1.13
This is higher than the required ratio, so Hg is the limiting reagent.
4.99 × 10-2 mol Hg ×
= 18.0 g HgBr2 produced
4.99 × 10-2 mol Hg ×
= 7.97 g Br2 reacted
excess Br2 = 9.00 g Br2 - 7.97 g Br2 = 1.03 g Br2
b. 5.00 mL Hg ×
= 0.339 mol Hg
5.00 mL Br2 ×
= 0.0970 mol Br2
Br2 is limiting since the actual moles of Br2 present is well below the required 1:1 mol ratio.
0.0970 mol Br2 ×
96.
1.50 g BaO2 ×
= 35.0 g HgBr2 produced
= 8.86 × 10-3 mol BaO2
= 1.87 × 10-2 mol HCl
25.0 mL ×
The required mol ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual ratio is:
= 2.11
Since the actual mol ratio is larger than the required mol ratio, the denominator (BaO2 ) is the limiting
reagent.
8.86 × 10-3 mol BaO2 ×
The amount of HCl reacted is:
= 0.301 g H2 O2
CHAPTER 3
STOICHIOMETRY
51
8.86 × 10-3 mol BaO2 ×
= 1.77 × 10-2 mol HCl
excess mol HCl = 1.87 × 10-2 mol - 1.77 × 10-2 mol = 1.0 × 10-3 mol HCl
mass of excess HCl = 1.0 × 10-3 mol HCl ×
97.
= 3.6 × 10-2 g HCl
Ca3 (PO4 )2 + 3 H2 SO4 ÷ 3 CaSO4 + 2 H3 PO4
1.0 × 103 g Ca3 (PO4 )2 ×
= 3.2 mol Ca3 (PO4 )2
1.0 × 103 g conc. H2 SO4 ×
= 10. mol H2 SO4
The required mol ratio from the balanced equation is 3 mol H2SO4 to 1 mol Ca3(PO4)2. The actual
ratio is:
= 3.1
This is higher than the required mol ratio, so Ca3 (PO4 )2 is the limiting reagent.
3.2 mol Ca3 (PO4 )2 ×
3.2 mol Ca3 (PO4 )2 ×
98.
= 1300 g CaSO4 produced
= 630 g H3 PO4 produced
An alternative method to solve limiting reagent problems is to assume each reactant is limiting and
calculate how much product could be produced from each reactant. The reactant that produces the
smallest amount of product will run out first and is the limiting reagent.
5.00 × 106 g NH3 ×
5.00 × 106 g O2 ×
5.00 × 106 g CH4 ×
= 2.94 × 105 mol HCN
= 1.04 × 105 mol HCN
= 3.12 × 105 mol HCN
O2 is limiting since it produces the smallest amount of HCN. Although more product could be
produced from NH3 and CH4 , only enough O2 is present to produce 1.04 × 105 mol HCN. The
mass of HCN produced is:
1.04 × 105 mol HCN ×
5.00 × 106 g O2 ×
= 2.81 × 106 g HCN
= 5.63 × 106 g H2O
52
99.
CHAPTER 3
STOICHIOMETRY
C2 H6 (g) + Cl2 (g) ÷ C2 H5 Cl(g) + HCl(g)
300. g C2 H6 ×
= 9.98 mol C2 H6; 650. g Cl2 ×
= 9.17 mol Cl2
The balanced equation requires a 1:1 mol ratio between reactants. 9.17 mol of C2 H6 will react with
all of the Cl2 present (9.17 mol). Since 9.98 mol C2H6 is present, Cl2 is the limiting reagent.
The theoretical yield of C2 H5 Cl is:
9.17 mol Cl2 ×
= 592 g C2 H5 Cl
Percent yield =
100.
× 100 = 82.8%
C7 H6 O3 + C4H6O3 ÷ C9H8O4 + HC2 H3 O2
1.50 g C7 H6 O3 ×
= 1.09 × 10-2 mol C7 H6 O3
2.00 g C4 H6 O3 ×
= 1.96 × 10-2 mol C4 H6 O3
C7 H6 O3 is the limiting reagent since the actual moles of C7 H6 O3 are below the required 1:1 mol ratio.
The theoretical yield of aspirin is:
1.09 × 10-2 mol C7 H6 O3 ×
% yield =
101.
= 1.96 g C9 H8 O4
× 100 = 76.5%
2.50 metric tons Cu3 FeS3 ×
= 1.39 × 106 g Cu (theoretical)
1.39 × 106 g Cu (theoretical) ×
= 1.20 × 106 g Cu = 1.20 × 103 kg Cu
= 1.20 metric tons Cu (actual)
102.
P4 (s) + 6 F2 (g) ÷ 4 PF3 (g); The theoretical yield of PF3 is:
120. g PF3 (actual) ×
= 154 g PF3 (theoretical)
CHAPTER 3
STOICHIOMETRY
154 g PF3 ×
53
×
×
= 99.8 g F2
99.8 g F2 are needed to produce an actual PF3 yield of 78.1%.
Additional Exercises
103.
The atomic mass is 54.94 amu. From the periodic table, the element is manganese (Mn).
104.
In one hour, the 1000. kg of wet cereal produced contains 580 kg H2 O and 420 kg of cereal. We want
the final product to contain 20.% H2 O. Let x = mass of H2 O in final product.
= 0.20, x = 84 + 0.20 x, x = 105 . 110 kg H2 O
The amount of water to be removed is 580 - 110 = 470 kg/hr.
105.
Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol; Since 104.14/13.02 = 7.998 . 8, the
molecular formula for styrene is (CH)8 = C8 H8 .
2.00 g C8 H8 ×
106.
×
41.98 mg CO2 ×
6.45 mg H2 O ×
×
= 11.46 mg C; %C =
= 0.722 mg H; %H =
= 9.25 × 1022 atoms H
× 100 = 57.85% C
× 100 = 3.64% H
%O = 100.00 - (57.85 + 3.64) = 38.51% O
Out of 100.00 g terephthalic acid, there are:
57.85 g C ×
= 4.817 mol C; 3.64 g H ×
38.51 g O ×
= 2.407 mol O
= 2.001;
= 1.50;
= 1.000
C:H:O mol ratio is 2:1.5:1 or 4:3:2. Empirical formula: C4 H3 O2
= 3.61 mol H
54
CHAPTER 3
STOICHIOMETRY
Mass of C4 H3 O2 . 4(12) + 3(1) + 2(16) = 83
Molar mass =
107.
17.3 g H ×
= 166 g/mol;
= 2; Molecular formula: C8H6O4
= 17.2 mol H; 82.7 g C ×
= 6.89 mol C
= 2.50; The empirical formula is C2H5.
The empirical formula mass is ~29 g, so two times the empirical formula would put the compound in
the correct range of the molar mass. Molecular formula = (C2 H5 )2 = C4 H10
2.59 × 1023 atoms H ×
= 4.30 × 10-2 mol C4 H10
4.30 × 10-2 mol C4 H10 ×
108.
= 2.50 g C4 H10
Assuming 100.00 g E3 H8 :
mol E = 8.73 g H ×
= 3.25 mol E
, x = molar mass of E = 28.1 g/mol; atomic mass of E = 28.1 amu
109.
Mass of H2 O = 0.755 g CuSO4CxH2O - 0.483 g CuSO4 = 0.272 g H2O
0.483 g CuSO4 ×
0.272 g H2 O ×
= 0.00303 mol CuSO4
= 0.0151 mol H2 O
; Compound formula = CuSO4C5H2O, x = 5
110.
a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer:
8.80 g N ×
% C3 H3N =
= 33.3 g C3 H3 N
= 33.3% C3H3N
CHAPTER 3
STOICHIOMETRY
55
Only butadiene in the polymer reacts with Br2:
0.605 g Br2 ×
= 0.205 g C4 H6
% C4 H6 =
× 100 = 17.1% C4H6
b. If we have 100.0 g of polymer:
33.3 g C3 H3 N ×
17.1 g C4 H6 ×
= 0.628 mol C3 H3 N
= 0.316 mol C4 H6
49.6 g C8 H8 ×
= 0.476 mol C8 H8
Dividing by 0.316:
This is close to a mol ratio of 4:2:3. Thus, there are 4 acrylonitrile to 2 butadiene to 3 styrene
molecules in the polymer or (A4 B2 S3 ).
111.
1.20 g CO2 ×
= 0.428 g C24 H30 N3 O
× 100 = 42.8% C24 H30 N3 O
112.
a. CH4 (g) + 4 S(s) ÷ CS2 (l) + 2 H2 S(g) or 2 CH4 (g) + S8 (s) ÷ 2 CS2 (l) + 4 H2 S(g)
b. 120. g CH4 ×
= 7.48 mol CH4; 120. g S ×
= 3.74 mol S
The required S to CH4 mol ratio is 4:1. The actual S to CH4 mol ratio is:
= 0.500
This is well below the required ratio so sulfur is the limiting reagent.
The theoretical yield of CS2 is: 3.74 mol S ×
= 71.2 g CS2
The same amount of CS2 would be produced using the balanced equation with S8 .
56
113.
CHAPTER 3
STOICHIOMETRY
453 g Fe ×
mass % Fe2 O3 =
114.
a. Mass of Zn in alloy = 0.0985 g ZnCl2 ×
%Zn =
= 0.0473 g Zn
× 100 = 9.34% Zn; %Cu = 100.00 - 9.34 = 90.66% Cu
b. The Cu remains unreacted. After filtering, washing, and drying, the mass of the unreacted copper
could be measured.
115.
Assuming one mol of vitamin A (286.4 g Vitamin A):
mol C = 286.4 g Vitamin A ×
= 20.00 mol C
mol H = 286.4 g Vitamin A ×
= 30.00 mol H
Since one mol of Vitamin A contains 20 mol C and 30 mol H, the molecular formula of Vitamin A
is C20 H30 E. To determine E, let’s calculate the molar mass of E.
286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/mol
From the periodic table, E = oxygen and the molecular formula of Vitamin A is C20H30O.
Challenge Problems
116.
; Assuming 100 atoms, let x = number of 85 Rb atoms and 100 - x = number of
atoms.
= 2.591, x = 259.1 - 2.591 x,
0.7215 (84.9117) + 0.2785 (A) = 85.4678, A =
117.
= 72.15%
85
87
Rb
Rb
= 86.92 amu = atomic mass of
First, we will determine composition in mass percent. We assume all the carbon in the 0.213 g CO2
came from 0.157 g of the compound and that all the hydrogen in the 0.0310 g H2 O came from the
0.157 g of the compound.
87
Rb
CHAPTER 3
STOICHIOMETRY
0.213 g CO2 ×
0.0310 g H2 O ×
57
= 0.0581 g C; %C =
× 100 = 37.0% C
= 3.47 × 10-3 g H; %H =
= 2.21% H
We get %N from the second experiment:
= 1.89 × 10-2 g N
0.0230 g NH3 ×
%N =
× 100 = 18.3% N
The mass percent of oxygen is obtained by difference:
%O = 100.00 - (37.0 + 2.21 + 18.3) = 42.5%
So out of 100.00 g of compound, there are:
37.0 g C ×
= 3.08 mol C; 2.21 g H ×
= 2.19 mol H
18.3 g N ×
= 1.31 mol N; 42.5 g O ×
= 2.66 mol O
The last, and often the hardest part, is to find simple whole number ratios. Divide all mole values by
the smallest number:
= 2.35;
= 1.67;
= 1.00;
= 2.03
Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6.
118.
1.0 × 106 kg HNO3 ×
= 1.6 × 107 mol HNO3
We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all 3
equations.
Thus, we can produce 16 mol HNO3 for every 24 mol NH3 we begin with:
1.6 × 107 mol HNO3 ×
= 4.1 × 108 g or 4.1 × 105 kg
This is an oversimplified answer. In practice, the NO produced in the third step is recycled back
58
CHAPTER 3
STOICHIOMETRY
continuously into the process in the second step. If this is taken into consideration, then the conversion
factor between mol NH3 and mol HNO3 turns out to be 1:1, i.e., 1 mol of NH3 produces 1 mol of
HNO3 . Taking into consideration that NO is recycled back gives an answer of
2.7 × 105 kg NH3 reacted.
119.
Total mass of copper used:
= 6.9 × 105 g Cu
10,000 boards ×
Amount of Cu removed = 0.80 × 6.9 × 105 g = 5.5 × 105 g Cu
5.5 × 105 g Cu ×
= 1.8 × 106 g Cu(NH3 )4 Cl2
5.5 × 105 g Cu ×
120.
= 5.9 × 105 g NH3
a. From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for every
4 mol of C6 H5 O3 N reacted. The actual yield is 3 moles of acetaminophen compared to a
theoretical yield of 4 moles of acetaminophen. Solving for percent yield by mass (where M =
molar mass acetaminophen):
% yield =
× 100 = 75%
b. The product of the percent yields of the individual steps must equal the overall yield, 75%.
(0.87) (0.98) (x) = 0.75, x = 0.88; Step III has a % yield = 88%.
121.
10.00 g XCl2 + excess Cl2 ÷ 12.55 g XCl4; 2.55 g Cl reacted with XCl2 to form XCl4. XCl4 contains
2.55 g Cl and 10.00 g XCl2 . From mol ratios, 10.00 g XCl2 must also contain 2.55 g Cl; mass X in
XCl2 = 10.00 - 2.55 = 7.45 g X.
2.55 g Cl ×
= 3.60 × 10-2 mol X
So, 3.60 × 10-2 mol X must equal 7.45 g X. The molar mass of X is:
; Atomic mass = 207 amu so X is Pb.
122.
4.000 g M2 S3 ÷ 3.723 g MO2
There must be twice as many mol of MO2 as mol of M2 S3 in order to balance M in the reaction.
Setting up an equation for 2 mol MO2 = mol M2 S3 where A = molar mass M:
CHAPTER 3
STOICHIOMETRY
59
2
,
8.000 A + 256.0 = 7.446 A + 358.2, 0.554 A = 102.2, A = 184 g/mol; atomic mass = 184 amu
123.
Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms O2- anions.
The simplest compound of the two elements is Al2 O3 . Similarly, we would expect the formula of any
group 6A element with Al to be Al2 X3 . Assuming this, out of 100.00 g of compound there are 18.56
g Al and 81.44 g of the unknown element, X. Let’s use this information to determine the molar mass
of X which will allow us to identify X from the periodic table.
18.56 g Al ×
= 1.032 mol X
81.44 g of X must contain 1.032 mol of X.
The molar mass of X =
= 78.91 g/mol X.
From the periodic table, the unknown element is selenium and the formula is Al2 Se3 .
124.
NaCl(aq) + Ag+ (aq) ÷ AgCl(s); KCl(aq) + Ag+ (aq) ÷ AgCl(s)
= 5.991 × 10-2 mol Cl-
8.5904 g AgCl ×
The molar masses of NaCl and KCl are 58.44 and 74.55 g/mol, respectively. Let x = g NaCl
and y = g KCl:
= 5.991 × 10-2 total mol Cl- or 1.276 x + y = 4.466
x + y = 4.000 g and
Solving using simultaneous equations:
1.276 x + y = 4.466
- x - y = -4.000
0.276 x
= 0.466,
% NaCl =
125.
x = 1.69 g NaCl and y = 2.31 g KCl
× 100 = 42.3% NaCl; % KCl = 57.7%
The balanced equations are:
4 NH3 (g) + 5 O2 (g) ÷ 4 NO(g) + 6 H2 O(g) and 4 NH3 (g) + 7 O2 (g) ÷ 4 NO2 (g) + 6 H2 O(g)
Let 4x = number of mol of NO formed, and let 4y = number of mol of NO2 formed. Then:
60
CHAPTER 3
STOICHIOMETRY
4x NH3 + 5x O2 ÷ 4x NO + 6x H2 O and 4y NH3 + 7y O2 ÷ 4y NO2 + 6y H2 O
All the NH3 reacted, so 4x + 4y = 2.00. 10.00 - 6.75 = 3.25 mol O2 reacted, so 5x + 7y = 3.25.
Solving by the method of simultaneous equations:
20 x + 28 y = 13.0
-20 x - 20 y = -10.0
8 y = 3.0, y = 0.38; 4x + 4 × 0.38 = 2.00, x = 0.12
mol NO = 4x = 4 × 0.12 = 0.48 mol NO formed