Download Ex. 38 Answer

New 21st Century Chemistry
Suggested answers to in-text activities and unit-end exercises
Topic 10 Unit 38
In-text activities
Checkpoint (page 85)
Gas
Mass of 1 mole of gas
Mass of 1 dm3 of gas
Molar volume
N2
28.0 g
1.166 g
24.0 dm3
O2
32.0 g
1.333 g
24.0 dm3
CO2
44.0 g
1.833 g
24.0 dm3
NH3
17.0 g
0.708 g
24.0 dm3
Checkpoint (page 88)
1
Gas
Mass of 1
mole of
gas
Mass of gas present
Number of moles of gas
Volume of gas at room
temperature and pressure
H2
2.0 g
0.40 g
0.20 mol
4.8 dm3
CO2
4.0 g
44.0 g
SO2
64.1 g
4.0 g
0.091 mol x 24.0 dm3 mol–1
=
0.091
mol
44.0 g mol–1
= 2.2 dm3
0.25 mol x 64.1 g
= 16 g
mol–1
1.24 mol x 32.0 g
= 39.7 g
mol–1
6 000 cm3
24 000 cm3 mol–1
6 000 cm3
= 0.25 mol
O2
2
32.0 g
1.24 mol
1.24 mol x 24.0 dm3 mol–1
= 29.8 dm3
Number of moles of methane gas
=
=
volume of methane gas (at room temperature and pressure)
molar volume of gas (at room temperature and pressure)
1.0 dm3
24.0 dm3 mol–1
= 0.042 mol
Suggested answers to in-text activities and unit-end exercises
Topic 10 Unit 38
1
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New 21st Century Chemistry
Number of methane molecules = number of moles of methane gas x L
= 0.042 mol x 6.02 x 1023 mol–1
= 2.5 x 1022
3
Number of moles of gas =
=
volume of gas (at room temperature and pressure)
molar volume of gas (at room temperature and pressure)
600 cm3
24 000 cm3 mol–1
= 0.0250 mol
Mass of gas = (78.98 –77.53) g
= 1.45 g
Molar mass of gas =
1.45 g
0.0250 mol
= 58.0 g mol–1
Checkpoint (page 92)
1 a) Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
b) As zinc was in excess, the amount of HCl used would limit the amount of H2 evolved.
Number of moles of HCl used = 0.250 mol dm–3 x
60.0
1 000
dm3
= 0.0150 mol
According to the equation, 2 moles of HCl react with Zn to produce 1 mole of H2.
i.e. number of moles of H2 =
0.0150
2
mol
= 0.00750 mol
Volume of gas evolved (at room temperature and pressure)
= number of moles of gas x molar volume of gas (at room temperature and pressure)
= 0.00750 mol x 24.0 dm3 mol–1
= 0.180 dm3
∴ 0.180 dm3 of gas would evolve in Experiment 1.
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Topic 10 Unit 38
2
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New 21st Century Chemistry
c)
2
a) No more gas was produced.
b) Calcium carbonate and nitric acid react according to the following equation:
CaCO3(s) + 2HNO3(aq)  Ca(NO3)2(aq) + H2O(l) + CO2(g)
Number of moles of gas collected (at room temperature and pressure)
=
=
volume of gas (at room temperature and pressure)
molar volume of gas (at room temperature and
pressure)
3
84.0 cm
24 000 cm3 mol–1
= 0.00350 mol
According to the equation, 1 mole of CaCO3 reacts with HNO3 to produce 1 mole of CO2.
i.e. number of moles of CaCO3 = 0.00350 mol
Mass of CaCO3 in the sample = 0.00350 mol x 100.1 g mol–1
= 0.350 g
Percentage by mass of CaCO3 in the sample =
0.350 g
0.380 g
x 100%
= 92.1%
c) The sample of calcite does not contain other carbonates that give carbon dioxide with nitric
acid.
Suggested answers to in-text activities and unit-end exercises
Topic 10 Unit 38
3
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New 21st Century Chemistry
Checkpoint (page 96)
1 CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
840 cm3 ? cm3
? cm3
Number of moles of CH4 =
=
volume of CH4 (at room temperature and pressure)
molar volume of gas (at room temperature and pressure)
840 cm3
24 000 cm3 mol–1
= 0.0350 mol
According to the equation, 1 mole of CH4 requires 2 moles of O2 for complete oxidation and
produces 1 mole of CO2.
a) Number of moles of O2 consumed = 2 x 0.0350 mol
= 0.0700 mol
Volume of O2 consumed (at room temperature and pressure)
= number of moles of O2 x molar volume of gas (at room temperature and pressure)
= 0.0700 mol x 24.0 dm3 mol–1
= 1.68 dm3
b) Number of moles of CO2 produced = 0.0350 mol
Volume of CO2 produced (at room temperature and pressure)
= number of moles of CO2 x molar volume of gas (at room temperature and pressure)
= 0.0350 mol x 24.0 dm3 mol–1
= 0.840 dm3
2
X(g) + 3Y(g)  2Z(g)
40 cm3
90 cm3
According to the equation, 1 mole of X reacts with 3 moles of Y to produce 2 moles of Z.
∴ 30 cm3 of X react with 90 cm3 of Y to produce 60 cm3 of Z, i.e. 10 cm3 of X would remain.
∴ volume of the resultant gaseous mixture = (10 + 60) cm3
= 70 cm3
Suggested answers to in-text activities and unit-end exercises
Topic 10 Unit 38
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New 21st Century Chemistry
Unit-end exercises (pages 98 – 104)
Answers for the HKCEE (Paper 1) and HKALE questions are not provided.
1
a) Number of moles of gas
=
volume of gas
molar volume of gas
b)
2
Formula
of gas
CH4
Mass of 1
mole of
gas
Number of moles
of gas present
Mass of gas present
4.0 g
16.0 g mol–1
4.00 g
16.0 g
Volume of gas at room
temperature and pressure
0.25 mol x 24.0 dm3 mol–1
= 6.0 dm
= 0.25 mol
NO2
46.0 g
3.00 dm3
24.0 dm3 mol–1
0.125 mol x 46.0 g mol–1
= 5.75 g
3.00 dm3
= 0.125 mol
NH3
Cl2
17.0 g
1.25 mol x 17.0 g mol–1
1.25 mol
= 21.3 g
3.55 g
71.0 g mol–1
3.55 g
71.0 g
1.25 mol x 24.0 dm3 mol–1
= 30.0 dm3
0.0500 mol x 24.0 dm3 mol–1
= 1.20 dm3
=0.0500 mol
3
C Number of moles of N2 =
7.00 g
28.0 g mol–1
= 0.250 mol
Molar mass of X =
15.0 g
0.250 mol
= 60.0 g mol–1
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Topic 10 Unit 38
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New 21st Century Chemistry
4
B 2Pb(NO3)2(s)  2PbO(s) + 4NO2(g) + O2(g)
0.30 mol
? dm3
According to the equation, 2 moles of Pb(NO3)2 decompose to give 1 mole of O2.
∴ number of moles of O2 obtained =
0.30
mol
2
= 0.15 mol
Volume of O2 obtained (at room temperature and pressure)
= number of moles of O2 x molar volume of gas (at room temperature and pressure)
= 0.15 mol x 24.0 dm3 mol–1
= 3.6 dm3
5
C
6
C 4H2(g) + Pb3O4(s)  3Pb(s) + 4H2O(l)
960 cm3
?g
Number of moles of H2 evolved
=
=
volume of H2 (at room temperature and pressure)
molar volume of gas (at room temperature and
pressure)
960 cm3
24 000 cm3 mol–1
= 0.0400 mol
According to the equation, 4 moles of H2 react with Pb3O4 to give 3 moles of Pb.
i.e. number of moles of Pb obtained =
3
4
x 0.0400 mol
= 0.0300 mol
Mass of Pb obtained = 0.0300 mol x 207.0 g mol–1
= 6.21 g
7
C XS(s) + O2(g)  X(l) + SO2(g)
2.91 g
300 cm3
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Topic 10 Unit 38
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New 21st Century Chemistry
Number of moles of SO2
=
=
volume of SO2 (at room temperature and pressure)
molar volume of gas (at room temperature and
pressure)
3
300 cm
24 000 cm3 mol–1
= 0.0125 mol
According to the equation, 1 mole of XS gives 1 mole of SO2 upon strong heating.
i.e. number of moles of XS = 0.0125 mol
Let m be the relative atomic mass of X
2.91 g
0.0125 mol
Molar mass of XS = (m + 32.1) g mol–1 =
m = 201
8
A 2NO(g) + O2(g)  2NO2(g)
1 dm3
2 dm3
? dm3
According to the equation, 2 moles of NO react with 1 mole of O2 to produce 2 moles of
NO2.
∴ 1 dm3 of NO reacts with 0.5 dm3 of O2 to produce 1 dm3 of NO2.
9
B
10 A
11 a) Density of CO2 =
=
mass (g)
volume (dm3)
0.7488 g
( 400.0 ) dm3
1 000
= 1.872 g dm–3
b) Molar mass of CO2 = 44.0 g mol–1
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Topic 10 Unit 38
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New 21st Century Chemistry
Volume of one mole of CO2 =
molar mass
density
44.0 g mol–1
=
1.872 g dm–3
= 23.5 dm3 mol–1
c) Number of moles of gas G
=
volume of gas G (under the experimental conditions)
molar volume of gas (under the experimental conditions)
200.0
) dm3
1 000
23.5 dm3 mol–1
(
=
= 8.51 x 10–3 mol
Molar mass of gas G =
0.5448 g
8.51 x 10–3 mol
= 64.0 g mol–1
12 a) 2CuS(s) + 3O2(g)  2CuO(s) + 2SO2(g)
1 910 g
?g
? dm3
b) Number of moles of CuS =
1 910 g
95.5 g mol–1
= 20.0 mol
According to the equation, 2 moles of CuS produce 2 moles of CuO and 2 moles of SO2.
i) Number of moles of CuO obtained = 20.0 mol
Mass of CuO obtained = 20.0 mol x 79.5 g mol–1
= 1 590 g
ii) Number of moles of SO2 given off = 20.0 mol
Volume of SO2 given off (at room temperature and pressure)
= number of moles of SO2 x molar volume of gas (at room temperature and pressure)
= 20.0 mol x 24.0 dm3 mol–1
= 480 dm3
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Topic 10 Unit 38
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New 21st Century Chemistry
13 a) Number of moles of N2
=
volume of N2 (under the conditions in the airbag)
molar volume of gas (under the conditions in the airbag)
54 dm3
=
20 dm3 mol–1
= 2.7 mol
b) 2NaN3(s)  2Na(l) + 3N2(g)
54 dm3
According to the equation, 2 moles of NaN3 decompose to produce 3 moles of N2.
i.e. number of moles of NaN3 =
2
3
x 2.7 mol
= 1.8 mol
c) Mass of NaN3 = 1.8 mol x 65.0 g mol–1
= 117 g
14 a) 4KO2(s) + 2H2O(g) + 4CO2(g)  4KHCO3(s) + 3O2(g)
?
7.68 dm3
Number of moles of CO2
=
=
volume of CO2 (at room temperature and pressure)
molar volume of gas (at room temperature and pressure)
7.68 dm3
24.0 dm3 mol–1
= 0.320 mol
According to the equation, 4 moles of KO2 react with 4 moles of CO2.
Assume the reaction is 100% efficient, then
number of moles of KO2 required = 0.320 mol
Mass of KO2 required = 0.320 mol x 71.1 g mol–1
= 22.8 g
b) The reaction is 80% efficient.
∴ mass of KO2 required = 22.8 g  80%
= 28.5 g
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Topic 10 Unit 38
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New 21st Century Chemistry
15 —
16 a)
b) The acid in flask B was more concentrated.
As zinc was in excess, the amount of HCl limited the amount of H2 evolved.
Hence the hydrochloric acid in flask B contained more HCl, i.e. the acid in flask B was more
concentrated.
c) Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
?g
60 cm3
Number of moles of H2 =
=
volume of H2 (at room temperature and pressure)
molar volume of gas (at room temperature and pressure)
60 cm3
24 000 cm3 mol–1
= 0.0025 mol
According to the equation, 1 mole of Zn reacts with HCl to produce 1 mole of H2.
∴ number of moles of Zn reacted = 0.0025 mol
Mass of Zn reacted = 0.0025 mol x 65.4 g mol–1
= 0.16 g
17 a) i) MgCO3(s) + 2HCl(aq)  MgCl2(aq) + CO2(g) + H2O(l)
ii) MgCO3(s) + 2H+(aq)  Mg2+(aq) + CO2(g) + H2O(l)
b) At the start, there are plenty of reactant particles per unit volume.
As the reactant particles are consumed gradually, there are fewer particles per unit volume.
So, the reaction slows down.
The reaction stops when one of the reactants is used up.
Suggested answers to in-text activities and unit-end exercises
Topic 10 Unit 38
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New 21st Century Chemistry
c) i)
ii) The rate of reaction would decrease.
Hydrochloric acid is a strong acid.
Hydrochloric acid almost completely dissociates in water while a weak acid only partially
dissociates.
Thus hydrochloric acid has a higher concentration of hydrogen ions than a weak acid of
the same concentration.
18 a) 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
b) Consider the production of CO2 from CaC2:
CaC2(s)  2CO2(g) (not a balanced chemical equation)
?g
10.8 dm3
Number of moles of CO2 =
=
volume of CO2 (at room temperature and pressure)
molar volume of gas (at room temperature and pressure)
10.8 dm3
24.0 dm3 mol–1
= 0.450 mol
According to the equation, 1 mole of CaC2 can produce 2 moles of CO2.
∴ number of moles of CaC2 reacted =
0.450
2
mol
= 0.225 mol
Mass of CaC2 reacted = 0.225 mol x 64.1 g mol–1
= 14.4 g
19 —
Suggested answers to in-text activities and unit-end exercises
Topic 10 Unit 38
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