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Chapter 4
Quantities in Chemical Reactions
Practice Problems
Problem 1
The difference in the activity of metals can be used to recover a metal from its compound
in a single displacement reaction such as the one shown below.
2Mg(s) + TiCl4(aq) → 2MgCl2(aq) + Ti(s)
(a) How many moles of Ti will be produced when 4.9 mol of Mg react completely?
(b) How many formula units of MgCl2 will be produced when 0.0582 mol of TiCl4 react
completely?
(c) How many atoms of Mg are required to react with 0.082 mol TiCl4?
Problem 2
How many atoms of lithium will produce 5.00 × 1028 molecules of H2(g) in the following
reaction?
2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)
Problem 3
A spectacular exothermic reaction occurs when ammonium dichromate is heated. The
reaction that occurs is shown by the following equation.
(NH4)2Cr2O7(s) → N2(g) + Cr2O3(s) + 4H2O(g) + heat
What mass of water vapour will be produced when 2.550 g of (NH4)2Cr2O7(s)
decompose?
Problem 4
What volume of H2(g) will be produced at STP when 4.28 g of calcium metal reacts with
water as shown by the following equation.
Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)
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Problem 5
Can 45.82 g of potassium chlorate generate 4.00 × 1023 molecules of oxygen gas in the
decomposition reaction that follows?
∆
→
2 KCl (s) + 3 O2 (g)
2 KClO3 (s) 
Problem 6
The fertilizer triple superphosphate, Ca(H2PO4)2 , can be produced by the following
reaction.
→ 3Ca(H2PO4)2 (aq)
Ca3(PO4)2 (s) + 4H3PO4 (aq) 
What mass of this fertilizer can be produced from 2.500 t of material that is 85.0%
calcium phosphate with an excess of phosphoric acid?
Problem 7
What mass of bismuth is required to produce 500.0 L of NO(g) at STP in the following
reaction?
Bi(s) + 4HNO3(aq) + 3H2O(l) → Bi(NO3)3(aq)·5H2O(aq) + NO(g)
Problem 8
The following single displacement reaction occurs when the solid mixture of aluminum
and manganese are heated.
2Al(s) + 3MnO(s) → Al2O3(s) + 3Mn(s)
What mass of manganese solid will be produced when 50.00 g of Al(s) and 90.00 g of
MnO(s) are allowed to react?
Problem 9
Calcium carbide is made in an electric furnace by the following reaction.
CaO(s) + 3C(s) → CaC2(s) + CO(g)
When 200.0 kg of CaO(s) and 300.0 kg of C(s) are reacted in the furnace, 182.4 kg of CaC2
are produced. What is the percentage yield in this reaction?
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Problem 10
When the copper ores mined near Sudbury are smelted, the following reaction occurs:
2Cu2O (s) + Cu2S (s) → 6Cu (s) + SO2 (g)
If this reaction proceeds 92.0% to completion, what mass of copper can be expected from
the smelting of a 1.00 t mixture that is 65.0 % Cu2O and 30.0 % Cu2S, with the remainder
being inert material?
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Answers
1.
n mol Ti
1 mol Ti
=
4.9 mol Mg 2 mol Mg
1 mol Ti
mol Ti = 4.9 mol Mg ×
= 2.4 mol Ti
2 mol Mg
(a)
2 mol MgCl 2
n mol MgCl 2
=
1 mol TiCl 4
0.0582 mol TiCl 4
2 mol MgCl 2
mol MgCl2 = 0.0582 molTiCl4 ×
= 0.116 mol MgCl2
1 mol TiCl 4
(b)
6.02 × 10 23 formula units
0.116 mol MgCl2 ×
= 6.98× 1022 formula units
1 mol
n mol Mg
2 mol Mg
=
0.082 mol TiCl 4 1 mol TiCl 4
2 mol Mg
mol of Mg = 0.082 mol TiCl4 ×
= 0.164 mol Mg
1 mol tiCl 4
(c)
0.164 mol Mg ×
6.02 × 10 23 atoms
= 9.87× 1022 atoms Mg
1 mol
2.
5.00× 1028 molecules H2(g) ×
1 mol
= 8.31× 104 mol H2(g)
6.02 × 10 23 molecules
n mol Li
2 mol Li
=
4
8.31 × 10 mol H 2(g) 1 mol H 2(g)
mol Li = n = 8.31× 104 mol H2(g) ×
1.66× 105 mol Li ×
2 mol Li
= 1.66× 105 mol Li
1 mol H 2(g)
6.02 × 10 23 atoms
= 9.99× 1028 atoms Li
1 mol
3.
molar mass (NH4)2Cr2O7(s) = 252.1 g/mol
2.550 g
mol (NH4)2Cr2O7(s) =
= 0.01012 mol
252.1 g/mol
4 mol H 2 O
n mol H 2 O
=
0.01012 mol (NH 4 ) 2 Cr2 O 7 1 mol (NH 4 ) 2 Cr2 O 7
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mol H2O = 0.01012 mol (NH4)2Cr2O7(s) ×
0.04048 mol H2O ×
4 mol H 2 O
= 0.04048 mol
1 mol (NH 4 ) 2 Cr2 O 7
18.02 g
= 0.7294 g
1 mol
4.
mol Ca(s) = n =
n mol H 2(g)
0.107 mol Li
=
m
4.28 g
=
= 0.107 mol Ca(s)
M 40.08 g/mol
1 mol H 2(g)
1 mol Li
mol H2(g) = n = 0.107 mol
at STP, 1 mol of any gas occupies 22.4 L
22.4 L
= 2.40 L at STP
volume of H2(g) = 0.107 mol ×
mol
5.
Molar mass of KClO3 = 122.55 g/mol
Moles of KClO3 = 45.82 g ×
3.739 × 10–1 mol KClO3 ×
5.608× 10–1 mol O2 ×
1 mol
= 3.739 × 10–1 mol KClO3
122.55 g
3 mol O 2
= 5.608× 10–1 mol O2
2 mol KClO 3
6.02 × 10 23 molecules
= 3.376× 1023 molecules O2
1 mol
This mass of potassium chlorate cannot generate 4.00× 1023 molecules of O2.
6.
Mass of Ca3(PO4)2 in starting material = 85.0% of 2.500 t = 2.125 t
Molar mass of Ca3(PO4)2 = 310.18 g/mol
Moles of Ca3(PO4)2 = 2.125 t × 1.000 × 106
6.851 × 103 mol Ca3(PO4)2 ×
g
1 mol
×
= 6.851 × 103 mol
t
310.18 g
3 mol Ca(H 2 PO 4 ) 2
= 2.055× 104 mol Ca(H2PO4)2
1 mol Ca 3 (PO 4 ) 2
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Molar mass of Ca(H2PO4)2 = 234.06 g/mol
Mass of Ca(H2PO4)2 = 2.055× 104 mol Ca(H2PO4)2 ×
= 4.810 × 106 g, or 4.810 t
234.06 g
=
1 mol
7.
1mol
= 22.3 mol
22.4 L
1 mol Bi
=
1 mol NO (g)
mol NO(g) = 500.0 L×
n mol Bi
22.3 mol NO (g)
mol Bi = n = 22.3 mol
molar mass Bi = M = 208.98 g/mol
mass Bi = n × M = 22.3 mol × 208.98 g/mol = 4.66 × 103 g Bi
8.
molar mass Al(s) = 26.98 g/mol molar mass MnO(s) = 70.94 g/mol
m
50.00 g
mol Al =
=
= 1.853 mol Al(s)
M 26.98 g/mol
mol MnO(s) =
m
90.00 g
=
= 1.269 mol MnO(s)
M 70.94 g/mol
Mn(s) produced by 1.853 mol Al(s) = 1.853 mol Al(s) ×
Mn(s) produced by 1.269 mol MnO(s) = 1.269 mol ×
3 mol Mn (s)
2 mol Al (s)
= 2.780 mol Mn(s)
3 mol Mn (s)
= 1.269 mol Mn(s)
3 mol MnO (s)
Since a smaller amount of Mn(s) is produced by the MnO(s), MnO(s) is the limiting reagent
and is used to determine the yield of Mn(s).
mass of Mn(s) = n × M = 1.269 mol × 54.94 g/mol = 69.72 g Mn
9.
molar mass CaO(s) = 56.08 g/mol
molar mass C(s) = 12.01 g/mol
molar mass CaC2(s) = 64.10 g/mol
m 200.0 kg ×1000 g/kg
=
= 3.566 × 103 mol CaO(s)
M
56.08 g/mol
m 300.0 kg ×1000 g/kg
=
= 2.498 × 104 mol C(s)
mol C(s) =
M
12.01 g/mol
mol CaO(s) =
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CaC2(s) produced from CaO(s) = 3.566 × 103 mol CaO(s) ×
1 mol CaC 2(s)
= 3.566 × 103 mol
1 mol CaO (s)
1 mol CaC 2(s)
= 8.327 × 103 mol
CaC2(s) produced from C(s) = 2.498 × 104 mol C(s) ×
3 mol C (s)
Since less CaC2(s) was produced from the CaO, CaO is the limiting reagent and the
theoretical yield of CaC2(s) will be based on this.
mass of CaC2(s) = n × M = 3.566 × 103 mol × 64.10 g/mol = 2.286 × 105 g
This is the theoretical yield. 2.286 × 105 g = 228.6 kg
percentage yield =
actual yield
182.4 kg
× 100% =
× 100% = 79.79 %
theoretical yield
228.6 kg
10.
Molar mass of Cu2O = 143.1 g/mol
Molar mass Cu2S = 159.1 g/mol
Moles of Cu2O = 65.0 % of 1.00 × 106 g ×
Moles of Cu2S = 30.0% of 1.00 × 106 g ×
1 mol
= 4.54 × 103 mol Cu2O
143.1 g
1 mol
= 1.89 × 103 mol Cu2S
159.1 g
Find the limiting reagent:
4.54 × 103 mol Cu2O produce 4.55 × 103 ×
6 mol Cu
2 mol Cu 2 O
1.89 × 103 mol Cu2S produce 1.89 × 103 mol ×
= 1.36 × 104 mol Cu
6 mol Cu
= 1.13 × 104 mol Cu
1 mol Cu S
2
Cu2S produces less copper, therefore Cu2S is the limiting reactant.
theoretical yield of Cu = 1.13 × 104 mol Cu ×
63.55 g copper
= 7.18 × 105 g
1 mol copper
0.92 × 7.18 × 105 g = 6.61 × 105 g
The actual yield will be 6.61 × 105 g.
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