Ch.4 Solutions
Solutions for Practice Problems p. 114
1. Consider the following reaction. 2H2(g) + O2(g) → 2H2O(l)
(a) Write the ratio of H2 molecules: O2 molecules: H2Omolecules.
The ratio is given by the whole number in front of each reactant and product in the
balanced equation. The ratio is 2: 1: 2
(b) How many molecules of O2 are required to react with 100 molecules of H2, according
to your ratio in part (a)
2 molecules H2: 1 molecule O2,
(half),therefore
100 molecules H2 : 50 molecules O2
(c) How many molecules of water are formed when 2478 molecules of O2 react with H2?
1 molecule O2 :2 molecules H2O,
(double), therefore
2478 molecules O2 : 4956 molecules H2O
(d) How many molecules of H2 are required to react completely with 6.02 x 1023
molecules of O2?
2 molecules H2 : 1 molecule O2
(double), therefore
1.20 x 1024 molecules H2: 6.02 x 1023 molecules O2
2. Iron reacts with chlorine gas to form iron(III) chloride, FeCl3.
2Fe(s) + 3Cl2(g) →
2FeCl3 (s)
(a) How many atoms of Fe are needed to react with three molecules of Cl2? 2 atoms
(b) How many formula units of FeCl3 are formed when 150 atoms of Fe react with
sufficient Cl2?
150 formula units (Ratio of 2:2)
(c) How many Cl2 molecules are needed to react with 1.204 x 1024 atoms of Fe?
1.204 x 1024 atoms Fe x 3 molecules Cl2 = 1.806 x 1024 molecules Cl2
2 Atoms Fe
( multiply value given by 3/2 ratio)
(d) How many formula units of FeCl3 are formed when 1.806 x 1024 molecules of Cl2
react with sufficient Fe?
1.806 x 1024 molecules of Cl2 x 2 formula units FeCl3 = 1.204 x 1024 formula units FeCl3
3 molecules Cl2
(multiply value given by ratio 2/3)
3. Consider the following reaction. Ca(OH)2(aq) + 2HCl(aq) → CaCl2 + 2H2O(l)
(a) How many formula units of calcium chloride, CaCl2, would be produced by 6.7 x 1025
molecules of hydrochloric acid, HCl?
2 molecules HCl : 1 formula unit CaCl2,
(divide by 2), therefore
25
25
6.7 x 10 molecules HCl : 3.35 x 10 formula units CaCl2
= 3.4 x 1025 formula units CaCl2
(b) How many molecules of water would be produced in the reaction in part (a)?
2 molecules HCl
:
2 molecules H2O,
(same ratio), therefore
25
25
6.7 x 10 molecules HCl : 6.7 x 10 molecules H2O
Solutions for Practice Problems p. 115
4. Aluminum bromide can be prepared by reacting small pieces of aluminum foil with
liquid bromine at room temperature. The reaction is accompanied by flashes of red light.
2Al(s) + 3Br2(l) → 2AlBr3(s)
How many moles of Br2 are needed to produce 5 mol of AlBr3, if sufficient Al is present?
5 mol AlBr3 x 3 mol Br2 = 7.5 mol Br2 = 8 mol Br2
2 mol AlBr3
5. Hydrogen cyanide gas, HCN(g), is used to prepare clear, hard plastics, such as
Plexiglas™. Hydrogen cyanide is formed by reacting ammonia, NH3, with oxygen and
methane, CH4.
2NH3 (g) + 3O2 (g) + 2CH4 (g) → 2HCN(g) + 6H2O(g)
(a) How many moles of O2 are needed to react with 1.2 mol of NH3?
(b) How many moles of H2O can be expected from the reaction of 12.5 mol of CH4?
Assume that sufficient NH3 and O2 are present.
(a) 1.2 mol of NH3 x 3 mol O2 = 1.8 mol O2
2 mol NH3
(b) 12.5 mol CH4 x 6 mol H2O = 37.5 mol H2O
2 mol CH4
6. Ethane gas, C2H6, is present in small amounts in natural gas. It undergoes complete
combustion to produce carbon dioxide and water.
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
(a) How many moles of O2 are required to react with 13.9 mol of C2H6?
13.9 mol C2H6 x 7 mol O2
= 48.65 mol O2 = 48.6 mol O2
2 mol C2H6
(b) How many moles of H2Owould be produced by 1.40 mol of O2 and sufficient ethane?
1.40 mol of O2 x 6 mol H2O = 1.20 mol O2
7 mol O2
7. Magnesium nitride reacts with water to produce magnesium hydroxide and ammonia
gas, NH3 according to the balanced chemical equation
Mg3N2 (s) + 6H2O(l) → 3Mg(OH)2 (s) + 2NH3 (g)
(a) How many molecules of water are required to react with 2.3 mol Mg3N2?
(b) How many formula units of Mg(OH)2 will be expected in part (a)?
(a) 2.3 mol Mg3N2 x 6 mol H2O = 13.8 mol H2O
1 mol Mg3N2
13.8 mol H2O x
6.02 x 1023 molecules H2O = 8.31 x 1024 molecules H2O
1 mol H2O
= 8.3 x 1024 molecules H2O
(b) 2.3 mol Mg3N2 x
3 mol Mg(OH)2
1 mol Mg3N2
= 6.9 mol Mg(OH)2
6.9 mol Mg(OH)2 x 6.02 x 1023 formula units Mg(OH)2 = 4.2 x 1024 fu Mg(OH)2
1 mol Mg(OH)2
Solutions for Practice Problems p. 125–126
19. Nitrogen gas is produced in an automobile air bag. It is generated by the
decomposition of sodium azide, NaN3.
2NaN3(s) → 3N2(g) + 2Na(s)
(a) To inflate the air bag on the driver’s side of a certain car, 80.0 g of N2 is required.
What mass of NaN3 is needed to produce 80.0 g of N2?
i) 80.0 g of N2 x 1 mol N2 = 2.855 mol N2
28.02 g N2
ii) 2.855 mol N2 x 2 mol NaN3 = 1.903 mol NaN3
3 mol N2
iii) 1.903 mol NaN3 x 65.02 g NaN3 = 124 g NaN3
1 mol NaN3
(b) How many atoms of Na are produced when 80.0 g of N2 are generated in this
reaction?
i) The number of moles of Na would be equivalent to the number of moles of NaN3 (ratio
of 2:2 in balanced equation). Therefore there would be 1.903 moles of Na produced.
ii) 1.903 mol Na x 6.02 x 1023 atoms Na
1 mol Na
= 1.1456 x 1024 = 1.14 x 1024 atoms Na
20. The reaction of iron(III) oxide with powered aluminum is known as the thermite
reaction.
2Al(s) + Fe2O3(s) → Al2O3 (s) + 2Fe(s)
(a) Calculate the mass of aluminum oxide, Al2O3, that is produced when 1.42 X 1024
atoms of Al react with Fe2O3.
1.42 X 1024 atoms Al x 1 mol Al = 2.358804 mol Al
6.02 x 1023 atoms
2.358804 mol Al x 1 mol Al2O3 =1.179402 mol Al2O3
2 mol Al
1.179402 mol Al2O3 x 101.96 g Al2O3 =120.3 g Al2O3 = 1.20 x 102 g Al2O3
1mol Al2O3
(b) How many formula units of Fe2O3 are needed to react with 0.134 g of Al?
0.134 g Al x 1 mol Al = 0.0049666 mol Al
26.98 g Al
0.0049666 mol Al x 1 mol Fe2O3= 0.0024833
2 mol Al
0.0024833 mol Fe2O3 x 6.02 x 1023 formula units Fe2O3
1 mol Fe2O3
21
= 1.49496 x 10 formula units Fe2O3
= 1.49 x 1021 formula units Fe2O3
Solutions for Practice Problems p. 131
23. The following balanced chemical equation shows the reaction of aluminum with
copper(II) chloride. If 0.25 g of aluminum reacts with 0.51 g of copper(II) chloride,
determine the limiting reactant.
2Al(s) + 3CuCl2(aq) → 3Cu(s) + 2AlCl3(aq)
Solution:
Reactant Al = 0.25 g
Reactant CuCl2 = 0.51 g
Convert the given masses into moles. From the balanced equation, identify the coefficient
that you will use to divide each quantity of moles, for comparison pursposes only (in
order to identify the limiting reactant (LR)
(i) 0.25 g Al x
1 mol Al
26.98 g Al
= 0.0092661 mol Al
0.0092661 mol Al /2 = 0.00463 mol Al
0.51 g CuCl2
x 1 mol CuCl2
=
[63.55 + 2(35.45)] g CuCl2
0.51mol CuCl2
134.45
= 0.0037932 mol CuCl2
0.0037932 mol CuCl2 / 3 = 0.0012644 mol CuCl2
The CuCl2 will run out first so it is the limiting reactant.
24. Hydrogen fluoride, HF, is a highly toxic gas. It is produced by the double
displacement reaction of calcium fluoride, CaF2, with concentrated sulfuric acid, H2SO4.
CaF2(s) + H2SO4(l) → 2HF(g) + CaSO4(s)
Determine the limiting reactant when 10.0 g of CaF2 reacts with 15.5 g of H2SO4.
Solution:
Reactant CaF2 = 10.0 g
Reactant H2SO4 = 15.5 g
(i) 10.0 g CaF2 x 1 mol CaF2
= 10.0 mol CaF2 = 0.12807 mol CaF2
[40.08 + 2(19.00)] g CaF2
78.08
15.5 g H2SO4
x
1 mol H2SO4
[2(1.01) + 32.07 + 4(16.00)] g H2SO4
= 15.5 mol H2SO4
98.09
= 0.15802 mol H2SO4
As the ratio of each reactant is 1:1, we can determine that CaF2 is the limiting reactant
(will run out first).
25. Acrylic, a common synthetic fibre, is formed from acrylonitrile C3H3N. Acrylonitrile
can be prepared by the reaction of propylene, C3H6, with nitric oxide, NO.
4C3H6 (g) + 6NO(g) → 4C3H3N(g) + 6H2O(g) + N2(g)
What is the limiting reactant when 126 g of C3H6 reacts with 175 g of NO?
Solution:
Reactant C3H6 = 126 g
Reactant NO = 175 g
(i) 126 g C3H6 x
1 mol C3H6
= 126 mol C3H6
[3(12.01 + 6(1.01)] g C3H6
42.09
175 g NO x
1 mol NO
=
(14.01 + 16.00) g NO
175 g
30.01 g/mol
= 2.9936 mol C3H6
= 5.8314 mol NO
(ii) Comparison of amount of moles (using mole ratio):
2.9936 mol C3H6 /4 = 0.7484 mol C3H6 (less so LR)
5.8314 mol NO /6 = 0.9719 mol NO
C3H6 is used up first, it is therefore the limiting reactant.
26. 3.76 g of zinc reacts with 8.93 x 1023 hydrogen ions as shown in the equation below.
Which reactant is present in excess?
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
Solution:
Reactant Zn = 3.76 g
Reactant H+ = 8.93 x 1023 ions.
NA = 6.02 x 1023 ions/mol
(i) 3.76 g Zn
x
8.93 x 1023 ions H+ x
1 mol Zn
65.39 g Zn
= 0.05750 mol Zn
1 mol H+ = 1.483 mol H+
6.02 x 1023 ions H+
(ii) 1.483 mol H+ / 2 = 0.7415 mol H+ (which is more than other reactant)
H+ is present in excess.
Solutions for Practice Problems p. 134–135
27. Chlorine dioxide, ClO2, is a reactive oxidizing agent. It is used to purify water.
6ClO2(g) + 3H2O(l) → 5HClO3(aq) + HCl(aq)
(a) If 71.00 g of ClO2 is mixed with 19.00 g of water, what is the limiting reactant?
(b) What mass of HClO3 expected in part (a)?
(c) How many molecules of HCl are expected in part (a)?
Given:
Mass of reactant ClO2 = 71.00 g
Mass of reactant H2O = 19.00 g
Solution:
(a) Determine the # moles of each reactant available, then divide that amount by the
coefficient in the balanced equation for comparative purposes in order to identify
the limiting reactant (LR).
71.00 g ClO2 x 1 mol ClO2
= 71.00 mol ClO2 = 1.052631579 mol ClO2
(35.45 + 32.00) g ClO2
67.45 g
1.052631579 mol ClO2 / 6 (coefficient) = 0.17544 mol ClO2
19.00 g H2O x 1 mol H2O
= 19.00 mol H2O = 1.054384018 mol H2O
(2.02 + 16.00) g H2O
18.02
1.054384018 mol H2O / 3 (coefficient) = 0.35146134 mol H2O
The limiting reactant is ClO2.
(b) Use the initial molar amount of the limiting reactant for future calculations.
1.052631579 mol ClO2 x 5 mol HClO3 = 0.87719 mol HClO3
6 mol ClO2
(0.87719 mol HClO3)(1.01 + 35.45 + 3x16.00) g = (0.87719)(84.46 g) = 74.09 g HClO3
1 mol
(c) Using the number of moles of ClO2 (the L.R.):
1.052631579 mol ClO2 x 1 mol HCl = 0.1754385965 mol HCl
6 mol ClO2
0.1754385965 mol HCl x 6.02 x 1023 molecules HCl = 1.056 x 1023 molecules HCl
mol HCl
28. Hydrazine, N2H4, reacts exothermically with hydrogen peroxide, H2O2.
N2H4(l) + 7H2O2(aq) → 2HNO3(g) + 8H2O(g)
(a) 120 g of N2H4 reacts with an equal mass of H2O2. Which is the limiting reactant?
(b) What mass of HNO3 is expected?
(c) What mass, in grams, of the excess reactant remains at the end of the reaction?
Solution:
N2H4(l) + 7H2O2(aq) → 2HNO3(g) + 8H2O(g)
120 g
120 g
(a) 120 g N2H4 x 1 mol N2H4
= 120 mol N2H4
(28.02 + 4.04) g N2H4
32.06
120 g H2O2 x 1 mol H2O2
=
(2.02 + 32.00) g H2O2
= 3.743 mol N2H4
120 mol H2O2 = 3.527 mol H2O2
34.02 g H2O2
The limiting reactant is H2O2. This answer may be supported in one of the following
ways:
i)
The number of moles of H2O2 is less than 7 times the amount it should be
as shown in the above balanced chemical equation. This means that H2O2
is used up first.
ii)
3.527 mol H2O2 x 1 mol N2H4 = 0.5038 mol N2H4
7 mol H2O2
A quantity which is smaller than the calculated availability of
N2H4.Therefore H2O2 is used up first
(b) Use LR (limiting reactant) for future calculations.
3.527 mol H2O2 x 2 mol HNO3 = 1.008 mol HNO3
7 mol H2O2
1.008 mol HNO3 x (1.01 + 14.01 + 3x16.00) g = (1.008)(63.02 g HNO3) = 63.52 g HNO3
1 mol HNO3
= 64 g HNO3
(c) Using the mole ratio of the limiting reactant, one can determine the amount of N2H4
that was used up. Subtract the amount used from the mole amount calculated in (a) for
the excess reactant. Multiply the mole difference by the molar mass of the excess reactant
to obtain the mass left over.
(i) 3.527 mol H2O2 x 1 mol N2H4 = 0.5039 mol N2H4 used up
7 mol H2O2
(ii)3.743 mol N2H4 available – 0.5039 mol N2H4 used = 3.239 mol N2H4 in excess
(iii) (3.239 mol N2H4) (2 x 14.01 + 4 x 1.01) g N2H4 = (3.239)(32.06 g N2H4)
1 mol N2H4
= 103.8 g N2H4 = 1.0 x 102 g N2H4 remains at the end of the reaction
29. In the textile industry, chlorine is used to bleach fabrics. Any of the toxic chlorine
that remains after the bleaching process is destroyed by reacting it with a sodium
thiosulfate solution, Na2S2O3(aq).
Na2S2O3(aq) + 4Cl2(g) + 5H2O(l) → 2NaHSO4(aq) + 8HCl(aq)
135 kg of Na2S2O3 reacts with 50.0 kg of Cl2 and 238 kg of water. How many grams of
NaHSO4 are expected to be produced?
What Is Given? Remember to convert the given kg quantities to grams before
proceeding with the calculations.
Na2S2O3(aq) + 4Cl2(g) + 5H2O(l) → 2NaHSO4(aq) + 8HCl(aq)
Reactant Na2S2O3 = 135 kg = 135 000 g
Reactant Cl2 = 50.0 kg = 5.00 x 104 g
Reactant H2O = 238 kg = 238 000 g
Step 1: Determine the limiting reactant:
(i) 135 000 g Na2S2O3 x 1 mol Na2S2O3
[2(22.99) + 2(32.07) + 3(16.00)] g
= 135 000 mol Na2S2O3
158.12
= 853.78 mol Na2S2O3
(For comparison purposes only so as to identify L.R., divide # mol by coefficient in
balanced equation above) 853.78 mol Na2S2O3 / 1 = 853.78 mol Na2S2O3
(ii) 5.00 x 104 g Cl2 x 1 mol Cl2
= 5.00 x 104 mol Cl2 = 705.22 mol Cl2
2 (35.45) g Cl2
70.90
(For comparison purposes only so as to identify L.R., divide # mol by coefficient in
balanced equation above) 705.22 mol Cl2 / 4 = 176.30 mol Cl2
(iii) 238 000 g H2O x 1 mol H2O
= 13208 mol H2O
18.02 g H2O
(For comparison purposes only so as to identify L.R., divide # mol by coefficient in
balanced equation above)
13208 mol H2O / 5
= 2641.6 mol H2O
The limiting reactant is chlorine (Cl2) as it is the reactant that will run out first.
[853.78 mol Na2S2O3 : 176.30 mol Cl2 : 2641.6 mol H2O]
Use the number of moles *available* of the LR for any remaining calculations.
Step 2: Cacu1ate the number of moles of NaHSO4 produced by the limiting reactant.
705.22 mol Cl2 x 2 mol NaHSO4 = 352.61 mol NaHSO4
4 mol Cl2
Step 3: Calculate the mass of NaHSO4
(352.61 mol NaHSO4)(22.99 + 1.01 + 32.07 + 64.00) g = (352.61)(120.07 g NaHSO4)
1 mol NaHSO4
= 42337.88 g NaHSO4 = 4.23 x 104 g NaHSO4
30. Manganese(III) fluoride can be formed by the reaction of manganese(II) iodide with
fluorine. 2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(l)
(a) 1.23 g of MnI2 reacts with 25.0 g of F2. What mass of MnF3 is expected?
(b) How many molecules of IF5 are produced in part (a)?
(c) What reactant is in excess? How much of it remains at the end of the reaction?
Solution:
Reactant MnI2 = 1.23 g
Reactant F2 = 25.0 g
(a) Determine the identity of the limiting reactant for the given amounts of reactants.
(i) 1.23 g MnI2 x 1 mol MnI2
= 1.23 mol MnI2 = 0.0039839 mol MnI2
(54.94 + 2 x 126.90) g
308.74
0.0039839 mol MnI2 /2 (coefficient) = 0.001992 mol MnI2
(ii) 25.0 g F2 x 1 mol F2
38.00
= 0.65789 mol F2
0.65789 mol F2 / 13 (coefficient) = 0.050607 mol F2
MnI2 is the limiting reactant as it will run out before F2
Use the original amount of moles of limiting reactant as starting point for all future
calculations.
(iii) 0.0039839 mol MnI2 x 2 mol MnF3 = 0.0039839 mol MnF3
2 mol MnI2
(0.0039839 mol MnF3)(54.94 + 3 x 19.00) g
1 mol MnF3
= (0.0039839 mol MnF3)(111.94 g/mol MnF3) = 0.44596 g = 0.446 g MnF3
(b) (i) 0.0039839 mol MnI2 x 4 mol IF5 = 0.0079678 mol IF5
2 mol MnI2
(ii) (0.0079678 mol IF5)(6.02 x 1023 molecules IF5) = 4.80 x 1021 molecules IF5
mol IF5
(c) Use the mole ratio of the limiting reactant to determine the number of moles of the
excess reactant, F2. Subtract this value from the initial calculated molar amount of F2.
(i) Calculate moles of excess reactant used up:
0.0039839 mol MnI2 x 13 mol F2 = 0.025895 mol F2 used
2 mol MnI2
(ii) 0.65789 mol F2 available - 0.025895 mol F2 used = 0.631995 mol F2 in excess
(iii) (0.631995 mol)(38.00 g/mol) = 24.0 g excess F2