AP Chemistry – Chapter 3 Assignment – Answers
1. a. The reaction can occur as long as the reactants are available in any amounts, without an exact
mole ratio you will have excess reactant.
c. A properly balanced reaction does allow you to know the amount of reactant(s) required to
produce a specific amount of product. C is a fair choice but D is better worded for intent.
e. A mole-to-mole ratio is not required for the reaction to occur as written. Excess reactants are
allowed and most reactions do not occur in whole number mole ratios.
4.
Only 4 atoms of nitrogen (from 2 diatomic molecules) are used in the production of ammonia before all
of the hydrogen is used up leaving excess molecules of nitrogen in the reaction vessel.
5. Choice d is the best choice because the reaction is balanced with the lowest whole number ratios.
Choice a shows what you start with and end up with in the reaction vessel but 4 of the molecules of
nitrogen are not reacting, they are staying the same therefore those nitrogen’s should not be part of the
equation for the actual reaction.
10.
a. F
b. F
c. F
d. T
e. F
11.
a. F
b. F
c. F
d. T
e. F
18.
a&c
22
a. Not enough information
28.
Proper formulas of the reactants and the products
b. T
c. F
d. F
e. F
Proper mole ratios between reactants and products
States of the reactants and products
In some cases, special reaction conditions, (Ex. Heat or catalyst added)
32. By determining the amount of product each reactant can make the limiting reactant is determined
to be the one that makes the least product. The other reactant must be in excess because, as in 4
above, once one of the reactants is used up the reaction stops.
38. Ag – 109
100% total – (51.82% Ag-107) = 48.18% Ag-109
Ag - 109
0.4818% x (atomic mass) = 52.470
Ag - 107
0.5182% x 106.905amu
= 55.398
TOTAL
= 107.868
Accepted mass
107.868 – 55.398 = 52.470
52.470/0.4818 = 108.904amu
44. 5.0 x 1021atoms C
1 mole C_____
6.02 x 1023 atoms C
8.3 x 10-3 mole C
12 g C________
1 mole C
= 8.3 x 10-3 mole C
= 0.10 g C
48. a. 219.88 g/mol
c. 141.96 g/mol
50. a. 4.55 x 10-3 mol P4O6
c. 7.04 x 10-3 mol Na2HPO4
52. a. 1.10 x 103 g Ca3(PO4)2
c. 7.10 x 102 g Na2HPO4
54. a. 619 g P
c. 155 g p
56. a. 2.74 x 1021 mc P4O6
c. 4,24 x 1921 f.u. Na2HPO4
58. a. 1.10 x 1022 atoms
c. 4.24 x 1021 atoms P
66. a. 74.09 g/mol
e. 1.2 x 10-16 g
c. 3.7 g
70. a. 49.47% C
c. 52.14% C
76. a. (SHN)4 or S4N4H4
c. Co2C8O8
82. empirical formula  PNCl2
molecular formula  P5N5Cl10
86. CH6N2
92. Fe3O4(s) + 4H2(g)  3Fe(s) + 4H2O(g)
Fe3O4(s) + 4CO(g)  3Fe(s) + 4CO2(g)
94. a. 2, 2, 2, 1, 1 or 4, 6, 4, 1, 4
c. 4, 5, 4, 6
e. 2, 5, 2, 1
g. 1, 1, 1
98. 1, 6, 1, 1, 3
100. 10, 3, 3, 10
36.8g P4O10
102. a. 1, 2, 1, 10,2
b. 3.2g NH4SCN
106. 68 hours or 2.8 days
110. 1, 3, 3, 2
1300g CaSO4
630g H3PO4
114. a. 1170g DDT
b. C2HOCl3 limiting, C6H5Cl excess
c. 401g C6H5Cl excess
120. 71.40%C
8.689% H
14.27% O
5.648% F
d. 17.1%
124. 32,000g or 32kg of bacterial tissue
128. 71.4g AgC10H9N4SO2 produced
Answers are given with proper Significant Digits for the most part and may vary slightly from yours.