unit 24 solution
DESCRIPTION
JK SOLUTIONTRANSCRIPT
Unit 24 Molar volume calculations
Suggested answers to in-text activities
Check Your Understanding (page 48)
1
Gas Mass of 1
mole of
gas
Mass of gas present Number of moles of gas Volume of gas of room
temperature and pressure
H2 2.0 g 0.4 g 0.20 mol 4.8 dm3
CO2 44.0 g 4.0 g 0.091 mol x 24.0 dm3 mol-1
= 2.2 dm3
SO2 64.0 g 0.25 mol x 64.0 g mol-1
= 16 g
6 000 cm3
O2 32.0 g 1.24 mol x 32.0 g mol-1
= 39.7 g
1.24 mol 1.24 mol x 24.0 dm3 mol-1
= 29.8 dm3
2 Number of moles of gas Y =
=
= 0.0250 mol
Molar mass of gas Y =
=
= 44.0 g mol-1
3 Number of moles of methane gas=
=
= 0.042 mol
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
Check Your Understanding (page 55)
1 Pb3O4(s) + 4H2(g) 3Pb(s) + 4H2O(l)
1 920 cm3 ? g
Number of moles of H2 =
=
= 0.0800 mol
According to the equation, 4 moles of H2 react to produce 3 moles of Pb.
∴ Number of moles of Pb produced = x 0.0800 mol
= 0.0600 mol
Mass of Pb produced = Number of moles of Pb x Molar mass of Pb
= 0.0600 mol x 207.0 g mol-1
= 12.4 g
2 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g)
0.60 mol ? cm3
According to the equation, 2 moles of ZnS react to produce 2 moles of SO2.
∴ Number of moles of SO2 produced = 0.60 mol
Volume of SO2 produced (at room temperature and pressure)
= Number of moles of SO2 x Molar volume of the gas (at room temperature and pressure)
= 0.60 mol x 24.0 dm3 mol-1
= 14.4 dm3
3 CaCO3(s) CaO(s) + CO2(g)
11.76 g ? dm3
Molar mass of CaO = (40.0 + 16.0) g mol-1
= 56.0 g mol-1
Number of moles of CaO =
=
= 0.210 mol
According to the equation, 1 mole of CaCO3 produces 1 mole of CaO and 1 mole of CO2 on
heating.
∴ Number of moles of CO2 given off = Number of moles of CaO formed
= 0.210 mol
Volume of CO2 given off (at room temperature and pressure)
= Number of moles of CO2 x Molar volume of the gas (at room temperature and
pressure)
= 0.210 mol x 24.0 dm3 mol-1
= 5.04 dm3
4 Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
0.96 g ? dm3
Number of moles of Mg =
=
= 0.040 mol
According to the equation, 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole
of H2.
∴ Number of moles of H2 formed = 0.040mol
Volume of H2 formed (at room temperature and pressure)
= Number of moles of H2 x Molar volume of the gas (at room temperature and
pressure)
= 0.040 mol x 24.0 dm3mol-1
= 0.96 dm3
5 NH3(g) + HNO3(aq) NH4NO3(aq)
36.0 dm3 ? g
Number of moles of NH3 =
=
= 1.50 mol
According to the equation, 1 mole of NH3 reacts with 1 mole of HNO3 to produce 1 mole of
NH4NO3.
∴ Number of moles of NH4NO3 produced = 1.50 mol
Molar mass of NH4NO3 = (2 x 14.0 + 4 x 1.0 + 3 x 16.0) g mol-1
= 80.0 g mol-1
Mass of NH4NO3 produced = Number of moles of NH4NO3 x Molar mass of NH4NO3
= 1.50 mol x 80.0 g mol-1
= 120 g
Check Your Understanding (page 58)
1 CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
840 cm3 ? cm3 ? cm3
Number of moles of CH4 =
=
= 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 the 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 the gas (at room temperature and
pressure)
= 0.0350 mol x 24.0 dm3mol-1
= 0.840 dm3
2 a) CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)
?g 2.00 M 72 cm3
120.0 cm3
b) All the egg shells have dissolved. The acid is in excess.
Number of moles of CO2 =
=
= 0.0030 mol
According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl to
produce 1 mole of CO2.
∴ Number of moles of CaCO3 reacted = 0.0030mol
Molar mass of CaCO3 = (40.0 + 12.0 + 3 x 16.0) mol-1
= 100.0 mol-1
Mass of CaCO3 reacted = Number of moles of CaCO3 x Molar mass of
CaCO3
= 0.0030 mol x 100.0 g mol-1
= 0.30 g
Percentage by mass of CaCO3 in the egg shells = x 100%
= 86%
Suggested answers to exercise
1
2
Formula
of gas
Mass of
1 mole
of gas
Mass of gas present Number of moles of gas
present
Volume of gas of room
temperature and
pressure
CH2 16.0 g 4.0 g 0.25 mol x 24.0 dm3 mol-1
= 6.0 dm3
NO2 46.0 g 0.125 mol x 46.0 g mol-
1
= 5.75 g
3.00 dm3
NH3 17.0 g 1.25 mol x 17.0 g mol-1
= 21.3 g
1.25 moles 1.25 mol x 24.0 dm3 mol-1
= 30.0 dm3
Cl2 71.0 g 3.55 g 0.0500 mol x 24.0 dm3
mol-1
= 1.20 dm3
3 D
Mass of a gas(in grams)
Number of moles of the gas
Volume of the gas at room temperature and
pressure (in dm3)
÷ molar mass
x molar mass ÷ 24.0 dm3 mol-1
x 24.0 dm3 mol-1
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 = x 0.30 mol
= 0.15 mol
Volume of O2 obtained (at room temperature and pressure)
= Number of moles of O2 x Molar volume of the gas (at room temperature and
pressure)
= 0.15 mol x 24.0 dm3 mol-1
= 3.6 dm3
5 A CaCO3(s) CaO(s) + CO2(g)
? g 960 cm3
Number of moles of CO2 =
=
= 0.0400 mol
According to the equation, 1 mole of CaCO3 produces 1 mole of CO2.
∴ Number of moles of CaCO3 required = 0.0400 mol
Molar mass of CaCO3 = 100.0 g mol-1
Mass of CaCO3 required = Number of moles of CaCO3 x Molar mass of CaCO3
= 0.0400 mol x 100.0 g mol-1
= 4.00 g
6 D CuO(s) + H2(g) Cu(s) + H2O(l)
? dm3 25.40 g
Number of moles of Cu =
=
= 0.400 mol
According to the equation, 1 mole of CuO reacts with 1 mole of H2 to produce 1 mole of
Cu.
∴ Number of moles of H2 required = 0.400 mol
Volume of H2 required (at room temperature and pressure)
= Number of moles of H2 x Molar volume of the gas (at room temperature and
pressure)
= 0.400 mol x 24.0 dm3 mol-1
= 9.60 dm3
7 C
8 B
9 A (3) Each hydrogen molecule contains two hydrogen atoms while helium is
monoatomic. Therefore, 1 mole of hydrogen gas and 1 mole of helium gas contain
different numbers of atoms.
10 B
11 a) Density of CO2 =
=
= 1.872 g dm-3
b) Molar mass of CO2 = 44.0 g mol-1
Volume of one mole of CO2 =
=
= 23.5 dm3 mol-1
c) Number of moles of gas G =
=
= 8.51 x 103 mol
Molar mass of gas G =
=
= 64.0 g mol-1
12 Mass of NaCl consumed per hour = (250 - 86) g dm-3 x 10 dm3
= 1 640 g
Molar mass of NaCl = (23.0 + 35.5) g mol-1
= 58.5 g mol-1
Number of moles of NaCl consumed per hour =
=
= 28.0 mol
According to the equation, 2 moles of NaCl produce 2 moles of NaOH and 1 mole of Cl2.
a) Number of moles of NaOH produced per hour = 28.0 mol
Molar mass of NaOH = (23.0 + 16.0 + 1.0) g mol-1
= 40.0 g mol-1
Mass of NaOH produced per hour = Number of moles of NaOH x Molar mass of NaOH
= 28.0 mol x 40.0 g mol-1
= 1 120 g
b) Number of moles of Cl2 produced per hour
= x 28.0 mol
= 14.0 mol
Volume of Cl2 produced per hour (at room temperature and pressure)
= Number of moles of Cl2 x Molar volume of the gas (at room temperature and
pressure)
= 14.0 mol x 24.0 dm3 mol-1
= 336 dm3
13 a) 2CuS(s) + 3O2(g) 2CuO(s) + 2SO2(g)
1 910 g ? g ? dm3
b) Molar mass of CuS = 95.5 g mol-1
Number of moles of CuS =
=
= 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
Molar mass of CuO = 79.5 g mol-1
Mass of CuO obtained= Number of moles of CuO x Molar mass of CuO
= 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 the gas (at room temperature and
pressure)
= 20.0 mol x 24.0 dm3 mol-1
= 480 dm3
14 a) 4KO2(s) + 2H2O(g) + 4CO2(g) 4KHCO3(s) + 3O2(g)
? g 7.68 dm3
Number of moles of CO2 =
=
= 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
Molar mass of KO2 = (39.0 + 2 x 16.0) g mol-1
= 71.0 g mol-1
Mass of KO2 required = Number of moles of KO2 x Molar mass of KO2
= 0.320 mol x 71.0 g mol-1
= 22.7 g
b) The reaction is 80% efficient.
∴ Mass of KO2 required = 22.7 g ÷ 80%
= 28.4 g
15 a) 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
b) We can represent the production process of CO2 by the following
equation:
CaC2(s) 2CO2(g) (not a balanced chemical equation)
? g 10.8 dm3
Number of moles of CO2 =
=
= 0.450 mol
According to the equation, 1 mole of CaC2 can produce 2 moles of CO2.
∴ Number of moles of CaC2 reacted = x 0.450 mol
= 0.225 mol
Molar mass of CaC2 = (40.0 + 2 x 12.0) g mol-1
= 64.0 g mol-1
Mass of CaC2 reacted = Number of moles of CaC2 x Molar mass of CaC2
= 0.225 mol x 64.0 g mol-1
= 14.4 g
16 2D2(g) + O2(g) 2D2O(l)
150 cm3 100 cm3
Number of moles of D2 =
=
= 0.00625 mol
Number of moles of O2 =
=
= 0.00417 mol
According to the equation, 2 moles of D2 react with 1 mole of O2 to produce 2
moles of D2O. O2 is in excess in this case. The amount of D2 limits the amount of D2O formed.
Number of moles of D2O produced = 0.00625 mol
Molar mass of D2O = (2 x 2.0 + 16.0) g mol-1
= 20.0 g mol-1
Mass of D2O produced = Number of moles of D2O x Molar mass of D2O
= 0.00625 mol x 20.0 g mol-1
= 0.125 g
17 NaHCO3(s) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)
0.420 g 114 cm3
Number of moles of CO2 =
=
= 0.00475 mol
According to the equation, 1 mole of NaHCO3 produces 1 mole of CO2.
∴ Number of moles of NaHCO3 reacted = 0.00475 mol
Molar mass of NaHCO3 = 84.0 g mol-1
Mass of NaHCO3 in the tablet = Number of moles of NaHCO3 x Molar mass
of NaHCO3
= 0.00475 mol x 84.0 g mol-1
= 0.399 g
Percentage purity of NaHCO3 in the tablet =
= 95.0%
18 a)
b) The acid in flask B is more concentrated.
c) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
? g 60 cm3
Number of moles of H2 =
=
= 0.0025 mol
According to the equation, 1 mole of Mg reacts with 2 moles of HCl to
produce 1 mole of H2.
∴ Number of moles of Mg reacted = 0.0025 mol
Mass of Mg reacted = Number of moles of Mg x Molar mass of Mg
= 0.0025 mol x 24.0 g mol-1
= 0.060 g