Workbook – Worked Solutions
Chapter 4 – The Periodic Table.....................................................75
Chapter 9 – The Mole Concept......................................................76
Chapter 10 – Properties of Gases ...................................................87
Chapter 11 – Stoichiometry ............................................................98
Chapter 13 – Volumetric Analysis: Acid-Base............................111
Chapter 15 – Volumetric Analysis: Oxidation-Reduction .........123
Chapter 16 – Rates of Reaction ....................................................135
Chapter 17 – Chemical Equilibrium............................................137
Chapter 18 – pH and Indicators...................................................144
Chapter 19 – Environmental Chemistry – Water.......................151
Chapter 21 – Fuels and Heats of Reaction...................................156
Chapter 24 – Stoichiometry II ......................................................160
Chemistry Live! – Worked Solutions
Workbook Chapter 4 – The Periodic Table
W4.2 (b)
In 100 atoms of boron there are:
81 atoms of mass 11 = 81 × 11 = 891
19 atoms of mass 10 = 19 × 10 = 190
–––––––––––––––––––––––––––––––––
Total mass of 100 atoms
= 1081
Average mass of 1 atom
= 10.81
Answer: Relative atomic mass of boron = 10.81
W4.3
In 100 atoms of neon there are
90 atoms of mass 20 = 90 × 20 = 1800
10 atoms of mass 22 = 10 × 22 = 220
–––––––––––––––––––––––––––––––––
Total mass of 100 atoms
= 2020
Average mass of 1 atom
= 20.2
Answer: Relative atomic mass of neon = 20.2
W4.5
In 100 atoms of nickel there are:
70 atoms of mass 58 = 70 × 58 = 4060
26 atoms of mass 60 = 26 × 60 = 1560
4 atoms of mass 62 = 4 × 62 = 248
––––––––––––––––––––––––––––––––––
Total mass of 100 atoms
= 5868
Average mass of 1 atom
= 58.68
Answer: Relative atomic mass of nickel = 58.68
75
Chemistry Live! – Worked Solutions
Workbook Chapter 9 – The Mole Concept
W9.1
(a) 1 mole of Li atoms = 7 g
(b) 1 mole of Na atoms = 23 g
(c)
1 mole of Ca atoms = 40 g
(d) 1 mole of Fe atoms = 56 g
(e) 1 mole of Ag atoms = 108 g
(f)
1 mole of Pb atoms = 207 g
W9.2
(a)
1 mole of Cl atoms
= 35.5 g
10 moles of Cl atoms = 35.5 × 10
= 355 g
(b)
1 mole of Br atoms
1 mole of Br2 molecules
0.125 moles of Br2 molecules
(c)
1 mole of H2 O = (1 × 2) + 16 = 18 g
0.25 mole of H2 O
= 18 × 0.25
= 4.5 g
(d)
1 mole of S atoms
1 mole of S8 molecules
0.5 moles of S8 molecules
(e)
1 mole of HNO3 molecules
= 1 + 14 + (16 × 3) = 63 g
0.5 moles of HNO3 molecules = 63 × 0.5
= 31.5 g
(f)
1 mole of SO4 2- ions
0.5 moles of SO4 2- ions
(g)
1 mole of O atoms
= 16 g
1 mole of O2 molecules
= 16 × 2 = 32 g
0.05 mole of O2 molecules = 32 × 0.05
= 1.6 g
76
= 80 g
= 80 × 2 = 160 g
= 160 × 0.125
= 20 g
= 32 g
= 32 × 8 = 256 g
= 256 × 0.5
= 128 g
= 32 + (16 × 4) = 96 g
= 96 × 0.5
= 48 g
Chemistry Live! – Worked Solutions
W9.3
(a)
Relative Molecular Mass NaOH
i.e 1 mole of NaOH
⇒ 4 moles of NaOH
= 23 + 16 + 1 = 40
= 40 g
= 40 × 4
= 160 g
= 40 + (2 × 35.5) = 40 + 71 = 111
= 111g
= 111 × 0.5
= 55.5 g
(b)
Relative Molecular Mass CaCl2
i.e 1 mole of CaCl2
⇒ 0.5 moles of CaCl2
(c)
Relative Molecular Mass Cl2
i.e. 1 mole of Cl2
⇒ 2 moles of Cl2
(d)
Relative Molecular Mass MgO
i.e. 1 mole of MgO
⇒ 0.05 mole of MgO
(e)
Relative Molecular Mass Fe2 O3
i.e. 1 mole of Fe2 O3
⇒ 3 moles of Fe2 O3
(f)
Relative Molecular Mass C2 H6
i.e. 1 mole of C2 H6
⇒ 0.1 mole C2 H6
(g)
Relative Molecular Mass CaCO3 = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100
i.e. 1 mole of CaCO3
= 100 g
⇒ 0.6 mole of CaCO3
= 100 × 0.6
= 60 g
(h)
Relative Molecular Mass NaCl = 23 + 35.5 = 58.5
i.e. 1 mole of NaCl
= 58.5 g
⇒ 0.3 moles of NaCl
= 58.5 × 0.3
= 17.55 g
(i)
Relative Molecular Mass H2 O = (2 × 1) + 16 = 18
i.e. 1 mole of H2 O
= 18 g
⇒ 4 moles of H2 O
= 18 × 4
= 72 g
= 2 × 35.5 = 71
= 71g
= 71 × 2
= 142 g
= 24 + 16 = 40
= 40 g
= 40 × 0.05
=2g
= (2 × 56) + (3 × 16) = 112 + 48 = 160
= 160 g
= 160 × 3
= 480 g
= (2 × 12) + (6 × 1) = 24 + 6 = 30
= 30 g
= 30 × 0.1
=3g
77
Chemistry Live! – Worked Solutions
(j)
Relative Molecular Mass Ca(NO3 )2
i.e.1 mole of Ca(NO3 )2
⇒ 0.075 moles of Ca (NO3 )2
= 40 + (14 × 2) + (16 × 6) = 40 + 28 + 96
= 164
= 164 g
= 164 × 0.075
= 12.3 g
(k)
Relative Molecular Mass (NH4 )2 SO4 = (2 × 14) + (8 × 1) + 32 + (4 × 16)
= 28 + 8 + 32 + 64
= 132
i.e. 1 mole of (NH4 )2 SO4 = 132 g
⇒ 0.02 moles of (NH4 )2 SO4 = 132 × 0.02
= 2.64 g
(l)
Relative Molecular Mass CuO
i.e. 1 mole of CuO
⇒ 0.25 moles of CuO
(m)
Relative Molecular Mass SO3
i.e. 1 mole of SO3
⇒ 0.1 moles of SO3
(n)
Relative Molecular Mass Cu2 O = (63.5 × 2) + 16 = 127 + 16 = 143
i.e.1 mole of Cu2 O
= 143 g
⇒ 0.25 moles of Cu2 O
= 143 × 0.25
= 35.75 g
(o)
Relative Molecular Mass (NH4 )2 CO3 = (2 × 14) + (8 × 1) + 12 + (3 × 16)
= 28 + 8 + 12 + 48
= 96
i.e. 1 mole of (NH4 )2 CO3 = 96 g
⇒ 3 moles of (NH4 )2 CO3 = 96 × 3
= 288 g
(p)
Relative Molecular Mass Br2
i.e. 1 mole of Br2
⇒ 0.25 moles of Br2
(q)
Relative Molecular Mass H2 SO4
i.e. 1 mole of H2 SO4
⇒ 0.8 moles of H2 SO4
78
= 63.5 + 16 = 79.5
= 79.5 g
= 79.5 × 0.25
= 19.875 g
= 32 + (3 × 16) = 32 + 48 = 80
= 80 g
= 0.1 × 80
=8g
= 2 × 80 = 160
= 160 g
= 160 × 0.25
= 40 g
= (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98
= 98 g
= 98 × 0.8
= 78.4 g
Chemistry Live! – Worked Solutions
(r)
Relative Molecular Mass FeCl2
i.e 1 mole of FeCl2
⇒ 4 moles of FeCl2
(s)
Relative Molecular Mass C12 H26
i.e. 1 mole of C12 H26
⇒ 1 × 10-4 moles of C12 H26
= 56 + (2 × 35.5) = 56 + 71 = 127
= 127 g
= 127 × 4
= 508 g
= (12 × 12) + (26 × 1) = 144 + 26 = 170
= 170 g
= 170 × 1 × 10-4
= 0.017 g
W9.4
(a)
Relative Molecular Mass CH4 = 12 + (4 × 1) = 12 + 4 = 16
Mass
80
Number of moles of CH4 = –––––––––––––––– = ––– = 5
Rel molecular mass
16
(b)
Relative Molecular Mass SO2 = 32 + (2 × 16) = 32 + 32 = 64
Mass
288
Number of moles of SO2 = –––––––––––––––– = –––– = 4.5
Rel molecular mass
64
(c)
Relative Molecular Mass H2 O = (2 × 1) + 16 = 2 + 16 = 18
Mass
0.32
Number of moles of H2 O = ––––––––––––––––– = ––––– = 0.0178
Rel molecular mass
18
(d)
Relative Molecular Mass H2 SO4 = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98
Mass
441
Number of moles of H2 SO4 = –––––––––––––––– = ––––– = 4.5
Rel molecular mass
98
(e)
Relative Molecular Mass NH4 NO3 = 14 + (4 × 1) + 14 + (3 × 16)
= 14 + 4 + 14 + 48
= 80
Mass
10
Number of moles of NH4 NO3 = –––––––––––––––– = ––– = 0.125
Rel molecular mass
80
79
Chemistry Live! – Worked Solutions
(f)
Relative Molecular Mass C6 H12 O6
= (6 × 12) + (12 × 1) + (6 × 16)
= 72 + 12 + 96
= 180
Mass
18
Number of moles of C6 H12 O6 = ––––––––––––––––– = –––– = 0.1
Rel molecular mass
180
W9.5
(a)
Relative Molecular Mass CaCO3 = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100
Mass
500
Number of moles of CaCO3 = ––––––––––––––––– = –––– = 5
Rel. molecular mass 100
(b)
Relative Molecular Mass NaCl = 23 + 35.5 = 58.5
Mass
351
Number of moles of NaCl = ––––––––––––––––– = –––– = 6
Rel. molecular mass 58.5
(c)
Relative Molecular Mass NaOH = 23 + 16 + 1 = 40
Mass
4
Number of moles of NaOH = ––––––––––––––––– = ––– = 0.1
Rel molecular mass 40
(d)
Relative Molecular Mass KNO3 = 39 + 14 + (3 × 16) = 39 + 14 + 48 = 101
Mass
25.25
Number of moles of KNO3 = ––––––––––––––––– = –––––– = 0.25
Rel molecular mass
101
(e)
Relative Molecular Mass CuSO4 .5H2 O
= 63.5 + 32 + (4 × 16) + (5 × 18)
= 63.5 + 32 + 64 + 90
= 249.5
Mass
24.95
Number of moles of CuSO4 .5H2 O = ––––––––––––––––– = –––––– = 0.1
Rel molecular mass
249.5
80
Chemistry Live! – Worked Solutions
W9.6
(a)
1 mole of H2 O contains 6 × 1023 molecules
0.25 moles of H2 O contain 0.25 × 6 × 1023 molecules
= 1.5 × 1023 molecules
(b)
1 mole of CH4 contains 6 × 1023 molecules
7.2 moles of CH4 contain 7.2 × 6 × 1023 molecules
= 4.32 × 1024 molecules
(c)
1 mole of CO2 contains 6 × 1023 molecules
8 moles of CO2 contain 8 × 6 × 1023 molecules
= 4.8 × 1024 molecules
(d)
1 mole of H2 SO4 contains 6 × 1023 molecules
0.001 mole of H2 SO4 contains 0.001 × 6 × 1023 molecules
= 6 × 1020 molecules
(e)
1 mole of NH3 contains 6 × 1023 molecules
5 × 10-3 mole of NH3 contains 5 × 10-3 × 6 × 1023
= 3 × 10 21 molecules
W9.7
(a)
Relative molecular mass O2 = 2 × 16 = 32
1 mole of O2 contains 6 × 1023 molecules
i.e. 32 g of O2 contain 6 × 1023 molecules
6 × 1023
1 g of O2 contains ––––––– molecules
32
6 × 1023 × 0.4
0.4 g of O2 contain ––––––––––– molecules
32
= 7.5 × 1021 molecules
(b)
Relative molecular mass Cl2 = 2 × 35.5 = 71
1 mole of Cl2 contains 6 × 1023 molecules
i.e. 71 g of Cl2 contain 6 × 1023 molecules
6 × 1023
1 g of Cl2 contains ––––––––––– molecules
71
6 × 1023 × 142
142 g of Cl2 contain ––––––––––––– molecules
71
= 1.2 × 1024 molecules
81
Chemistry Live! – Worked Solutions
(c)
Relative molecular mass CO2 = 12 + (2 × 16) = 44
1 mole of CO2 contains 6 × 1023 molecules
i.e. 44g of CO2 contain 6 × 1023 molecules
6 × 1023
1 g of CO2 contains ––––––––– molecules
44
40 × 6 × 1023
40 g of CO2 contain –––––––––––– molecules
44
= 5.45 × 1023 molecules
(d)
Relative molecular mass CH4 = 12 + (4 × 1) = 16
1 mole of CH4 contains 6 × 1023 molecules
i.e.16 g of CH4 contain 6 × 1023 molecules
6 × 1023
1 g of CH4 contains –––––––– molecules
16
50 × 6 × 1023
50 g of CH4 contain –––––––––––– molecules
16
= 1.875 × 1024 molecules
W9.8
(a)
Relative atomic mass Al = 27
i.e. 1 mole of Al = 27 g
⇒ 6 × 1023 atoms of Al = 27 g
27
1 atom of Al = –––––––– g
6 × 1023
2 × 10
25
27 × 2 × 1025
atoms of Al = –––––––––––– g
6 × 1023
= 900 g
82
Chemistry Live! – Worked Solutions
(b)
Relative atomic mass Mg = 24
i.e. 1 mole of Mg = 24 g
⇒ 6 × 1023 atoms of Mg = 24 g
24
1 atom of Mg = –––––––– g
6 × 1023
24 × 3 × 1021
3 × 1021 atoms of Al = –––––––––––– g
6 × 1023
= 0.12 g
(c)
Relative atomic mass Ag = 108
i.e. 1 mole of Ag = 108 g
⇒ 6 × 1023 atoms of Ag = 108 g
108
1 atom of Ag = ––––––– g
6 × 1023
108 × 1.15 × 1012
1.15 × 1012 atoms of Ag = –––––––––––––––– g
6 × 1023
= 2.07 × 10-10 g
(d)
Relative molecular mass H2 O = (1 × 2) + 16 = 2 + 16 = 18
i.e. 1 mole of H2 O = 18 g
⇒ 6 × 1023 molecules of H2 O = 18 g
18
1 molecule of H2 O = –––––––– g
6 × 1023
18 × 1.5 × 1025
1.5 × 1025 molecules of H2 O = ––––––––––––– g
6 × 1023
= 450 g
83
Chemistry Live! – Worked Solutions
(e)
Relative molecular mass CO2 = 12 + (2 × 16) = 12 + 32 = 44
i.e. 1 mole of CO2 = 44 g
⇒ 6 × 1023 molecules of CO2 = 44 g
44
1 molecule of CO2 = ––––––– g
6 × 1023
8 × 10
21
44 × 8 × 1021
molecules of CO2 = ––––––––––– g
6 × 1023
= 0.5867 g
Revision Question W9.9
(b)
1 mole of Al2 (SO4 )3 contains 2 moles of Al3+
⇒ 3 moles of Al2 (SO4 )3 contain 2 × 3 = 6 moles Al3+
(c)
Relative Molecular Mass Na2 SO3
= (2 × 23) + 32 + (3 × 16)
= 46 + 32 + 48
= 126
1 mole of Na2 SO3 contains 6 × 1023 SO3 2- ions
i.e. 126 g of Na2 SO3 contain 6 × 1023 SO3 2- ions
6 × 1023
1 g of Na2 SO3 contains –––––––– SO3 2- ions
126
2.1 × 6 × 1023
2.1 g of Na2 SO3 contain –––––––––––– SO3 2- ions
126
= 1 × 1022 SO3 2- ions
(d)
1 mole of Mg contain 6 × 1023 atoms
i.e. 24 g of Mg contain 6 × 1023 atoms
6 × 1023
1 g of Mg contains –––––––– atoms
24
0.36 × 6 × 1023
0.36 g Mg contains –––––––––––––– atoms
24
= 9 × 1021 atoms
84
Chemistry Live! – Worked Solutions
(e)
1 mole of Ca contains 6 × 1023 atoms of Ca
i.e. 40g of Ca contain 6 × 1023 atoms of Ca
6 × 10 23
1g Ca contains –––––––– atoms of Ca
40
8 × 6 × 1023
8g Ca contain ––––––––––– atoms of Ca
40
= 1.2 × 1023 atoms of Ca
6 × 1023 atoms of Mg = 24g
24
1 atom of Mg = –––––––– g
6 × 1023
24 × 1.2 × 1023
1.2 × 1023 atoms of Mg = –––––––––––– g
6 × 1023
= 4.8g
(f)
Mass
0.2
Number of moles of Mg = ––––––––––––––––––– = –––– = 8.33 × 10–3 mole
Relative Atomic Mass
24
(g)
Mass
7
Number of moles of Fe = –––––––––––––––––– = ––– = 0.125 mole
Relative Atomic Mass 56
0.125 mole of Mg will contain the same number of atoms as 0.125 mole of Fe.
1 mole of Mg = 24 g
⇒ 0.125 mole Mg = 0.125 × 24 = 3 g
(h) Relative molecular mass HNO3 = 1 + 14 + (3 × 16) = 63
Mass
6.3
Number of moles of HNO3 = –––––––––––––––– = –––– = 0.1 mole
Rel molecular mass 63
1 mole of HNO3 contains 3 moles of O atoms
⇒ 0.1 mole of HNO3 contain 3 × 0.1
= 0.3 mole of O atoms.
85
Chemistry Live! – Worked Solutions
(i)
Relative molecular mass CH4 = 12 + (4 × 1) = 16
Mass
6.4
Number of moles of CH4 = ––––––––––––––––– = –––– = 0.4
Rel molecular mass
16
1 mole of CH4 contains 4 moles of H atoms
⇒ 0.4 mole of CH4 contain 4 × 0.4
= 1.6 moles of H atoms
(j)
Mass
22
Number of moles of Cu = –––––––––––––––––– = ––––– = 0.346
Relative Atomic Mass 63.5
Mass
46
Number of moles of Al = –––––––––––––––––– = –––– = 1.704
Relative Atomic Mass 27
Mass
2.6
Number of moles of Fe = –––––––––––––––––– = –––– = 0.046
Relative Atomic Mass 56
(k)
Mass
39
Number of moles of K = –––––––––––––––––– = ––– = 1
Relative Atomic Mass 39
1 mole of K will contain the same number of atoms as 1 mole of C.
Mass of 1 mole of C = 12 g
-
(l)
1 mole of Mg(NO3 )2 contains 2 moles of NO3 ions
⇒ 2 moles of Mg(NO3 )2 contain 2 × 2
= 4 moles of NO3 - ions
(m)
Relative molecular mass CO2 = 12 + (2 × 16) = 44
Mass
22
Number of moles of CO2 = –––––––––––––––– = ––– = 0.5
Rel molecular mass
44
0.5 mole of CO2 will contain the same number of molecules as 0.5 mole of H2 O.
1 mole of water
⇒ 0.5 mole of water = 0.5 × 18 = 9 g
86
= 18 g
Chemistry Live! – Worked Solutions
Workbook Chapter 10 - Properties of Gases
W10.1
Given
V1 = 30 cm3
p1 = 975 kPa
T1 = 25 °C + 273 = 298 K
p1 V1
–––––
T1
s.t.p.
V2 = ?
p2 =100 kPa
T2 = 273 K
p2 V2
–––––
T2
=
975 × 30
–––––––– =
298
100 × V2
––––––––
273
⇒ V2
975 × 30 × 273
= ––––––––––––––
298 × 100
= 267.96 cm3
W10.2
Given
V1 = 2.5 L
p1 = 300 kPa
T1 = T2
p1 V1
s.t.p.
V2 = ?
p2 =100 kPa
p2 V2
–––––
=
–––––
T1
T2
300 × 2.5 =
100 × V2
300 × 2.5
⇒ V2 = ––––––––––
100
= 7.5 L
87
Chemistry Live! – Worked Solutions
W10.3
Old
V1 = 1.2 m3
p1 = 101 kPa
T1 = 5 °C + 273 = 278 K
New
V2 = ?
p2 = 108 kPa
T2 = 30 °C + 273 = 303 K
.
p1 V1
p2 V2
–––––
=
–––––
T1
T2
101 × 1.2
––––––––––
278
108 × V2
= ––––––––––
303
101 × 1.2 × 303
⇒ V2 = ––––––––––––––––
278 × 108
= 1.22 m3
W10.5
(a)
Rel molecular mass O2 = 2 × 16 = 32
i.e. 1 mole of O2 = 32 g
32 g of O2 occupies 22.4 L at s.t.p
22.4
1 g of O2 occupies ––––– L
32
22.4 × 10
⇒10 g of O2 occupies –––––––––– L
32
=7L
(b)
Rel molecular mass CO2 = 12 + (2 × 16) = 12 + 32 = 44
i.e. 1 mole of CO2 = 44 g
44 g of CO2 occupies 22.4 L at s.t.p
22.4
1 g of CO2 occupies ––––– L
44
22.4 × 12
⇒12 g of CO2 occupies –––––––––– L
44
= 6.11 L
88
Chemistry Live! – Worked Solutions
(c)
Rel molecular mass NH3 = 14 + (3 × 1) = 14 + 3 = 17
i.e. 1 mole of NH3 = 17 g
17 g of NH3 occupies 22.4 L at s.t.p
22.4
1 g of NH3 occupies –––––– L
17
22.4 × 8
⇒ 8 g of NH3 occupies –––––––––– L
17
= 10. 54 L
(d)
Rel molecular mass CO = 12 + 16 = 28
i.e. 1 mole of CO = 28 g
28 g of CO occupies 22.4 L at s.t.p
22.4
1 g of CO occupies –––––– L
28
22.4 × 25
⇒ 25 g of CO occupies –––––––––– L
28
= 20 L
(e)
Rel molecular mass NO2 = 14 + (2 × 16) = 14 + 32 = 46
i.e. 1 mole of NO2 = 46 g
46 g of NO2 occupies 22.4 L at s.t.p
22.4
1 g of NO2 occupies –––––– L
46
22.4 × 12
⇒ 12 g of NO2 occupies –––––––––– L
46
= 5.84 L
89
Chemistry Live! – Worked Solutions
Rel molecular mass H2 S = (2 × 1) + 32 = 2 + 32 = 34
i.e. 1 mole of H2 S = 34 g
34 g of H2 S occupies 22.4 L at s.t.p
(f)
22.4
1 g of H2 S occupies ––––– L
34
22.4 × 13
⇒ 13 g of H2 S occupies –––––––––– L
34
= 8.56 L
W10.6
p = 200 kPa = 200 × 103 Pa
V = 100 cm3 = 100 × 10-6 m3
n=?
R = 8.31 J mol-1 K-1
T = 30 °C + 273 = 303 K
pV = nRT
pV
n = ––––
RT
200 × 103 × 100 × 10-6
= ––––––––––––––––––––––
8.31 × 303
= 7.94 × 10-3 moles
90
Chemistry Live! – Worked Solutions
W10.7
p = 101.325 kPa = 101325 Pa
V = 512 cm3 = 512 × 10-6 m3
n=?
R = 8.31 J mol-1 K-1
T = 20 °C + 273 = 293 K
Mass of liquid = 1.236 g
pV = nRT
pV
n = ––––
RT
101325 × 512 × 10-6
= –––––––––––––––––––––
8.31 × 293
= 0.0213 mole
i.e. 0.0213 mole = 1.236 g
1.236
1 mole of liquid = ––––––
0.0213
= 58.03
i.e. Relative molecular mass of liquid = 58 (to nearest whole number)
91
Chemistry Live! – Worked Solutions
W10.8
p = 101.3 kPa = 101300 Pa
V = 200 cm3 = 200 × 10-6 m3
n=?
R= 8.31 J mol-1 K-1
T= 25 °C + 273 = 298 K
Mass of liquid = 1.0 g
pV = nRT
pV
n = –––
RT
101300 × 200 × 10-6
= ––––––––––––––––––––
8.31 × 298
= 8.18 × 10-3 mole
i.e. 8.18 × 10-3 mole = 1.0 g
1.0
1 mole of liquid = ––––––––––––
8.18 × 10-3
= 122.25
i.e. Relative molecular mass of gas = 122 (to nearest whole number)
92
Chemistry Live! – Worked Solutions
W10.9
Laboratory
V1 = 328 cm3
p1 = 9.8 × 104 Pa
T1 = 100 °C + 273 = 373 K
p 1 V1
s.t.p.
V2 = ?
p2 = 1 × 105 Pa
T2 = 273 K
p2 V2
=
–––––
–––––
T1
T2
9.8 × 104 × 328
1 × 105 × V2
––––––––––––––––
373
= –––––––––––––––
273
9.8 × 104 × 328 × 273
⇒ V2 = –––––––––––––––––––––––
373 × 1 × 105
= 235.26 cm3
Mass of condensed liquid = 102.84 – 101.41 = 1.43 g
At s.t.p. 235.26 cm3 of the vapour have a mass of 1.43g
1.43
3
1 cm of the vapour has a mass of ––––––– g
235.26
22,400 × 1.43
⇒ 22,400 cm of the vapour have a mass of –––––––––––––––
235.26
3
= 136.16 g
i.e Relative molecular mass = 136 (to nearest whole number)
93
Chemistry Live! – Worked Solutions
W10.10
p = 1.02 × 105 Nm-2
V = 67 – 6 = 61 cm3 = 61 × 10-6 m3
n=?
R = 8.31 J mol-1 K-1
T = 100 °C + 273 = 373 K
Mass of liquid injected = 13.179 – 12.910 = 0.269 g
pV = nRT
pV
n = –––
RT
1.02 × 105 × 61 × 10-6
= –––––––––––––––––––––––
8.31 × 373
= 2.01 × 10-3 mole
i.e. 2.01 × 10-3 mole = 0.269 g
0.269
1 mole of liquid = ––––––––––––
2.01 × 10-3
= 133.83
i.e. Relative molecular mass of liquid = 134 (to nearest whole number)
94
Chemistry Live! – Worked Solutions
W10.11
p = 1 × 105 Nm-2
V = 4.98 × 10-3 m3
n=?
R = 8.31 J mol-1 K-1
T = 27 °C + 273 = 300 K
pV = nRT
pV
n = ––––
RT
1 × 105 × 4.98 × 10-3
= ––––––––––––––––––––––
8.31 × 300
= 0.20 moles
i.e. 0.20 moles = 5.6 g
5.6
1 mole of gas = –––––
0.2
= 28
i.e. Relative molecular mass of diatomic gas = 28
⇒ gas = N2
The element is nitrogen.
95
Chemistry Live! – Worked Solutions
W10.12
(iii)
p = 1 × 105 Nm-2
V = 3.15 × 10-4 m3
n=?
R = 8.31 J mol-1 K-1
T = 300 K
pV = nRT
pV
n = –––
R
1 × 105 × 3.15 × 10-4
= ––––––––––––––––––––––
8.31 × 300
= 1.26 × 10-2 mole
i.e. 1.26 × 10-2 mole = 1.1 g
1.1
1 mole of gas = –––––––––––
1.26 × 10-2
= 87.30
i.e. Relative molecular mass of gas = 87.30
W10.13
(b)
s.t.p
V1 = 500 cm3
p1 = 1.01 × 105 Nm-2
T1 = 273 K
p 1 V1
new
V2 = ?
p2 = 2.02 × 105 Nm-2
T2 = 819 K
p2 V2
=
–––––
–––––
T1
T2
1.01 × 105 × 500
2.02 × 105 × V2
–––––––––––––––––
=
–––––––––––––––
273
819
1.01 × 105 × 500 × 819
⇒ V2 = ––––––––––––––––––––––––
2.02 × 105 × 273
= 750 cm3
96
Chemistry Live! – Worked Solutions
(c)
Rel molecular mass N2 = 2 × 14 = 28
i.e. 1 mole of N2 = 28 g
28 g of N2 occupy 22.4 L at s.t.p.
22.4
⇒1 g of N2 occupies ––––– L = 0.8 L
28
(e)
Mass
Density = –––––––––
Volume
2.5
Mass of one mole
Mass of one mole
= –––––––––––––––––––– = –––––––––––––––––––
Molar volume
22.4
⇒ Mass of one mole = 22.4 × 2.5 = 56 g
i.e. Relative molecular mass of gas = 56
W10.14
(vi)
p = 1.0 × 105 Nm-2
V = 168 cm3 = 168 × 10-6 m3
n=?
R = 8.4 N m mol-1 K-1
T = 300 K
pV = nRT
pV
n = –––––
RT
1.0 × 105 × 168 × 10-6
= ––––––––––––––––––––––
8.4 × 300
= 6.67 × 10-3 moles
i.e. 6.67 × 10-3 moles = 0.3 g
0.3
1 mole of liquid = ––––––––––––
6.67 × 10-3
= 44.98
i.e. Relative molecular mass of liquid = 44.98
97
Chemistry Live! – Worked Solutions
Workbook Chapter 11 - Stoichiometry
W11.1
(a)
Relative molecular mass Na2 CO3 .10H2 O
= (23 × 2) + 12 + (16 × 3) + 10 [(1 × 2) + 16]
= 46 + 12 + 48 + (10 × 18)
= 46 +12 + 48 + 180 = 286
180
% H2 O = ––––– × 100 = 62.94 %
286
(b)
Relative molecular mass MgSO4 .7H2 O
= 24 + 32 + (16 × 4) + 7 [(1 × 2) + 16]
= 24 + 32 + 64 + (7 × 18)
= 24 + 32 + 64 + 126 = 246
126
% H2 O = ––––– × 100 = 51.22 %
246
W11.2
Relative molecular mass Fe2 O3
= (56 × 2) + (16 × 3)
= 112 + 48 = 160
112
% Fe = ––––– × 100 = 70 %
160
Relative molecular mass Fe3 O4 = (56 × 3) + (16 × 4)
= 168 + 64 = 232
168
% Fe = ––––– × 100 = 72.41%
232
Magnitite Fe3 O4 contains the higher % of Fe.
W11.3
Element
Percentage
Percentage
–––––––––––––––
Carbon
Hydrogen
Oxygen
98
Rel. At. Mass
64.9
64.9
–––––– = 5.41
12
13.5
13.5
––––– = 13.5
1
21.6
21.6
–––––– = 1.35
16
i.e. Empirical formula = C4 H10 O
Simplest ratio
(divide by 1.35)
5.41
–––––
=4
1.35
13.5
–––––
= 10
1.35
1.35
–––––
1.35
=1
Chemistry Live! – Worked Solutions
Molecular Formula = empirical formula × n
74 = (C 4 H10 O) × n
74 = [(12 × 4) + (1 × 10) + 16] n
74 = (48 + 10 + 16) n
74 = 74 n
n=1
⇒ Molecular Formula = (C 4 H10 O) × 1
= C4 H10 O
i.e. Molecular Formula = empirical formula = C4 H10 O
W11.4
Element
Percentage
Percentage
–––––––––––––––
Carbon
26.7
Hydrogen
Oxygen
2.2
71.1
Rel. At. Mass
26.7
––––– = 2.23
12
2.2
–––– = 2.2
1
71.1
––––– = 4.44
16
Simplest ratio
(divide by 2.2)
2.23
–––––
=1
2.2
2.2
–––––
=1
2.2
4.44
–––––
=2
2.2
i.e. Empirical formula = CHO2
Molecular Formula = empirical formula × n
90 = (CHO 2 ) × n
90 = [12 + 1 + (16 × 2)] n
90 = (12 + 1 + 32) n
90 = 45n
n=2
⇒ Molecular Formula = (CHO 2 ) × 2
= C2 H2 O4
i.e. Molecular Formula = C2 H2 O4
99
Chemistry Live! – Worked Solutions
W11.5
Element
Percentage
Percentage
–––––––––––––––
Carbon
64.8
Hydrogen
Oxygen
13.6
21.6
Rel. At. Mass
64.8
––––– = 5.4
12
13.6
––––– = 13.6
1
21.6
–––––– = 1.35
16
i.e. Empirical formula = C4 H10 O
Molecular Formula = empirical formula × n
74 = (C 4 H10 O) × n
74 = [(12 × 4) + (1 × 10) + 16] n
74 = (48 + 10 + 16) n
74 = 74n
n=1
⇒ Molecular Formula = (C 4 H10 O) × 1
= C4 H10 O
i.e. Molecular Formula = C4 H10 O
100
Simplest ratio
(divide by 1.35)
5.4
––––––
=4
1.35
13.6
–––––
= 10
1.35
1.35
–––––
1.35
=1
Chemistry Live! – Worked Solutions
W11.6
Lead + iodine → lead iodide
2.07 g
4.61 g
mass of lead = 2.07 g
mass of lead iodide = 4.61 g
⇒ mass of iodine in lead iodide = 4.61 – 2.07 = 2.54 g
mass
number of moles of Pb in lead iodide = –––––––––––––––––
Rel. atomic mass
2.07
= –––––– = 0.01
207
mass
number of moles of I in lead iodide = –––––––––––––––––
Rel. atomic mass
2.54
= ––––– = 0.02
127
Ratio of Pb : I = 0.01 : 0.02 = 1 : 2
⇒ Empirical Formula = PbI2
101
Chemistry Live! – Worked Solutions
W11.7
CuSO4 .xH2 O → CuSO4 + x H2 O
3.94 g
2.52 g
mass of CuSO4 = 2.52 g
mass of CuSO4 .xH2 O = 3.94 g
mass of H2 O in CuSO4 .xH2 O = 3.94 – 2.52 = 1.42 g
Rel. mol. mass CuSO4 = 63.5 + 32 + (16 × 4)
= 63.5 + 32 + 64 = 159.5
Rel. mol. mass H2 O = (1 × 2) + 16 = 18
mass
number of moles of CuSO4 in CuSO4 .xH2 O = ––––––––––––––
Rel. mol. mass
2.52
= –––––– = 0.0158
159.5
mass
number of moles of water in CuSO4 .xH2 O = ––––––––––––––––
Rel. mol. mass
1.42
= ––––– = 0.0789
18
Ratio of CuSO4 : H2 O = 0.0158 : 0.0789 = 1 : 5
i.e. CuSO4 .5H2 O is the formula of the hydrated salt.
102
Chemistry Live! – Worked Solutions
W11.8
M
+ oxygen → M2 O
9.76 g
20.9 g
mass of M = 9.76 g
mass of M2 O = 20.9 g
mass of O2 = 20.9 – 9.76 = 11.14 g
mass
number of moles of O2 = –––––––––––––––
Rel. mol. mass
11.14
= ––––––– = 0.6963
16
Since in the compound M2 O the ratio M : O = 2 : 1
⇒ number of moles of metal = 0.6963 × 2 = 1.3926
i.e. 1.3926 moles of M = 9.76 g
9.76
1 mole of M = ––––––––
1.3926
= 7.01 ≈ 7
i.e. Relative atomic mass of M = 7
W11.9
Rel. mol. mass Mg3 N2 = (24 × 3) + (14 × 2)
= 72 + 28 = 100
i.e.
3Mg + N2 →
3Mg
→
3 moles →
3 (24) g →
72 g Mg →
1 g Mg
Mg3 N2
Mg3 N2 (shortened version of equation)
1 mole
100 g
100 g
100
→ –––– g Mg3 N2
72
33 × 100
33 g Mg → –––––––– g Mg3 N2
72
= 45.83 g Mg3 N2
103
Chemistry Live! – Worked Solutions
W11.10
Relative molecular mass Fe2 O3 = (56 × 2) + (16 × 3)
= 112 + 48 = 160
2Al + Fe2 O3
Fe2 O3
1 mole
160 g
160 g Fe2 O3
→
→
→
→
→
1 g Fe2 O3
112
→ –––– g Fe
160
30 g Fe2 O3
30 × 112
→  g Fe
160
Al2 O3 + 2Fe
2Fe (shortened version of equation)
2 moles
2 (56) g
112 g Fe
= 21 g Fe
W11.11
Relative mol. mass NH4 NO3 = 14 + (1 × 4) + 14 + (16 ×3)
= 14 + 4 +14 + 48 = 80
NH4 NO3
NH4 NO3
1 mole
i.e. 80 g NH4 NO3
80 g NH4 NO3
80
–––––– g NH4 NO3
22,400
→
→
→
→
→
N2 O + 2H2 O
N2 O (shortened version of equation)
1 mole
22.4 L of N2 O at s.t.p.
22,400 cm3 of N2 O
→ 1cm3 of N2 O
550 × 80
 g NH4 NO3 → 550 cm3 of N2 O
22,400
= 1.96 g NH4 NO3
104
Chemistry Live! – Worked Solutions
W11.12
Rel. molecular mass CaC2 = 40 + (12 × 2)
= 40 + 24 = 64
CaC2 + 2H2 O
CaC2
1 mole
i.e. 64 g CaC2
64 g CaC2
64
 g CaC2
22,400
→
→
→
→
→
C2 H2 + Ca(OH)2
C2 H2 (shortened version of equation)
1 mole
22.4 L of C2 H2 at s.t.p.
22,400 cm3 of C2 H2
→ 1cm3 of C2 H2
350 × 64
 g CaC2 → 350cm3 of C2 H2
22,400
= 1 g CaC2
W11.13
Relative mol. mass NH4 Cl = 14 + (1 × 4) + 35.5
= 14 + 4 + 35.5 = 53.5
2NH4 Cl + Ca(OH)2
2NH4 Cl
2 moles
2 (53.5) g
i.e. 107 g NH4 Cl
107 g NH4 Cl
107
 g NH4 Cl
44,800
→ 2NH3 + CaCl2 + 2H2 O
→ 2NH3 (shortened version of equation)
→ 2 moles
→ 2 (22.4) L of NH3 at s.t.p.
→ 44.8 L of NH3
→ 44,800 cm3 of NH3
→1cm3 of NH3
107 × 800
 g NH4 Cl → 800 cm3 of NH3
44,800
= 1.91 g NH4 Cl
105
Chemistry Live! – Worked Solutions
W11.14
Relative mol. mass KClO 3 = 39 + 35.5 + (16 × 3)
= 39 + 35.5 + 48 = 122.5
2KClO 3
2KClO 3
2 moles
2 (122.5) g
i.e. 245 g KClO3
245 g KClO 3
→
→
→
→
→
→
2KCl + 3O2
3O2 (shortened version of equation)
3 moles
3 (22.4) L of O2 at s.t.p
67.2 L of O2
67,200 cm3 of O2
245
 g KClO 3 → 1cm3 of O2
67,200
700 × 245
 g KClO 3 → 700 cm3 of O2
67,200
= 2.55 g KCl
W11.15
Mass
number of moles of Cu =       
Rel. atomic mass
12.7
=   = 0.2
63.5
mass
number of moles of S =       
Rel. atomic mass
3.2
=  = 0.1
32
(iii)
Ratio of Cu : S = 0.2 : 0.1 = 2 : 1
(iv)
Cu2 S
copper (I) sulfate
106
Chemistry Live! – Worked Solutions
W11.16
Relative mol. mass CaCO3 = 40 + 12 + (16 × 3)
= 40 + 12 + 48
= 100
2CaCO3 + 2SO2 + O2 → 2CaSO4 + 2CO2
2 moles of CaCO3 reacts with 2 moles of SO2
1 mole of CaCO3 reacts with 1 mole of SO2
100 g CaCO3 reacts with 22.4 L of SO2
22.4
1 g CaCO3 reacts with  L of SO2
100
1000 × 22.4
1000g CaCO3 reacts with  L of SO2
100
= 224 L of SO2
W11.17
Na2 SO3 + 2HCl → 2NaCl + SO2 + H2 O
(i)
Relative mol. mass Na2 SO3 = (23 × 2) + 32 + (16 × 3)
= 46 + 32 + 48 = 126
mass
number of moles of Na2 SO3 = 
Rel. mol. Mass
6.3
=  = 0.05
126
(ii)
Na2 SO3 → SO2 (shortened version of equation)
1 mole → 1 mole
1 mole → 22.4 L
0.05 mole of Na2 SO3 → (0.05 × 22.4) L of SO2 at s.t.p.
= 1.12 L of SO2
(iii)
1 mole of SO2 contains 6 ×1023 molecules
22.4 L of SO2 contains 6 × 1023 molecules
6 × 1023
1 L of SO2 contains  molecules
22.4
1.12 × 6 × 1023
1.12 L of SO2 contains  molecules
22.4
= 3 × 1022 molecules
107
Chemistry Live! – Worked Solutions
W11.18
4FeS2 + 11O2    → 2Fe2 O3 + 8SO2
(i)
Rel. mol. mass FeS2 = 56 + (32 × 2)
= 56 + 64 = 120
Rel. mol. mass Fe2 O3 = (56 × 2) + (16 × 3)
= 112 + 48 = 160
i.e.
4FeS2
4 moles
4(120) g
480 g FeS2
→
→
→
→
1 g FeS2
320
→   g Fe2 O3
480
9.6 g FeS2 →
2Fe2 O3 (shortened version of equation)
2 moles
2(160) g
320 g Fe2 O3
9.6 × 320

480
= 6.4 g Fe2 O3
(ii)
4FeS2
4 moles
4(120) g
480 g FeS2
→ 8SO2 (shortened version of equation)
→ 8 moles
→ 8 (22.4) L
→ 179.2 L of SO2 at s.t.p.
1 g FeS2
179.2
→ 

 L of SO2
480
9.6 × 179.2
9.6 g FeS2 → 



 of SO2
480
= 3.584 L of SO2
108
Chemistry Live! – Worked Solutions
1 mole of SO2 contains 6 ×1023 molecules
22.4 L of SO2 contains 6 × 1023 molecules
(iii)
6 × 1023
1 L of SO2 contains  molecules
22.4
3.584 × 6 × 1023
3.584 L of SO2 contains  molecules
22.4
= 9.6 × 1022 molecules
W11.19
(a)
Element
Percentage
Carbon
38.7
Hydrogen
9.68
Oxygen
100 – (38.7 + 9.68)
= 100 – 48.38
= 51.62
Percentage
–––––––––––––
Rel. At. Mass
38.7
––––– = 3.23
12
9.68
––––– = 9.68
1
Simplest ratio
(divide by 3.23)
3.23
–––– = 1
3.23
9.68
–––– = 3
3.23
51.62
––––– = 3.23
16
3.23
–––– = 1
3.23
i.e. Empirical formula = CH3 O
(b)
Molecular Formula = empirical formula × n
62 = (CH3 O) × n
62 = [12 + (1 × 3) + 16] n
62 = (12 + 3 + 16) n
62 = 31n
n=2
⇒ Molecular Formula = (CH3 O) × 2
= C2 H6 O2
i.e. Molecular Formula = C2 H6 O2
109
Chemistry Live! – Worked Solutions
W11.20
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2 O
(i)
Rel. mol. mass MnO2 = 55 + (16 × 2)
= 55 + 32 = 87
mass
number of moles of MnO2 = –––––––––––––
Rel. mol. mass
4.35
= –––– = 0.05
87
(ii)
1 mole of MnO2 reacts with 4 moles of HCl
⇒ 0.05 mole of MnO 2 reacts with (0.05 × 4) mole of HCl
= 0.2 mole of HCl
(iii)
Rel. mol. mass MnCl2 = 55 + (35.5 × 2)
= 55 + 71 = 126
MnO2 → MnCl2 (shortened version of equation)
1 mole → 1 mole
0.05 mole → 0.05 mole
Mass of MnCl2 = Number of moles × Rel. mol. mass
= 0.05 × 126
= 6.3 g
(iv)
MnO2 → Cl2 (shortened version of equation)
1 mole → 1 mole
1 mole → 22.4 L
0.05 mole MnO 2 → (0.05 × 22.4) L of Cl2 at s.t.p.
= 1.12 L of Cl2
(v)
1 mole of Cl2 contains 6 × 1023 molecules
0.05 mole of Cl2 contains 0.05 × 6 ×1023 molecules
= 3 ×1022 molecules
110
Chemistry Live! – Worked Solutions
Workbook - Chapter 13 – Volumetric Analysis: Acid - Base
W 13.1
37.9 % w/v HCl means there are 37.9 g HCl in 100 g of solution
⇒
100
1 g HCl in    g of solution
37.9
5 × 100
5 g HCl in    g of solution
37.9
= 13.19 g
W13.2
(a) 0.54 g/L = 0.54 × 1000 mg/L
= 540 mg/L
= 540 ppm
(b) 0.18 g/L = 0.18 × 1000 mg/L
= 180 mg/L
= 180 ppm
(c) 0.077 g/100 cm3
= 0.077 × 10 g/L
= 0.77 g/L
= 0.77 × 1000 mg/L
= 770 mg/L
= 770 ppm
(d) 0.0009 g/100 cm3 = 0.0009 × 10 g/L
= 0.009 g/L
= 0.009 × 1000 mg/L
= 9 mg/L
= 9 ppm
111
Chemistry Live! – Worked Solutions
W13.3
(a) Rel. mol mass HCl = 1 + 35.5 = 36.5
36.5 g HCl in 1L of solution → 1M solution
1
1g HCl in 1L of solution →  M solution
36.5
65
65 g HCl in 1L of solution →  M solution
36.5
= 1.78 M
(b) Rel. mol mass KOH = 39 + 16 + 1 = 56
56 g KOH in 1L of solution → 1M solution
1
1 g KOH in 1L of solution →  M solution
56
25
25 g KOH in 1L of solution →  M solution
56
25 × 4
25 g KOH in 250 cm of solution →  M solution
56
3
= 1.79 M
(c) Rel. mol mass H2 SO4 = 2(1) + 32 + 4(16)
= 2 + 32 + 64 = 98
98 g H2 SO4 in 1L of solution → 1M solution
1
1g H2 SO4 in 1L of solution →  M solution
98
22
22 g H2 SO4 in 1L of solution →  M solution
98
22 × 10
22 g H2 SO4 in 100 cm of solution →  M solution
98
3
= 2.24 M
112
Chemistry Live! – Worked Solutions
(d) Rel. mol mass NaOH = 23 + 16 + 1 = 40
40 g NaOH in 1L of solution → 1M solution
1
1 g NaOH in 1L of solution →  M solution
40
10
10 g NaOH in 1L of solution →  M solution
40
10
10g NaOH in 2 L of solution →  M solution
40 × 2
= 0.125 M
(e) Rel. mol mass Na2 CO3 = 2(23) + 12 + 3(16)
= 46 + 12 + 48 = 106
106 g Na2 CO3 in 1L of solution → 1M solution
1
1 g Na2 CO3 in 1L of solution →  M solution
106
12.5
12.5 g Na2 CO3 in 1L of solution →  M solution
106
12.5 × 5
12.5 g Na2 CO3 in 200 cm3 of solution →  M solution
106
= 0.59 M
113
Chemistry Live! – Worked Solutions
W13.4
80 mg /100 cm3 = 0.08 g /100 cm3
Rel. mol mass C2 H5 OH = 2(12) + 5(1) + 16 +1
= 24 + 5 + 16 + 1 = 46
46 g C2 H5 OH in 1L of solution → 1M solution
1
1 g C2 H5 OH in 1L of solution →  M solution
46
0.08
0.08 g C2 H5 OH in 1L of solution →  M solution
46
0.08 × 10
0.08 g C2 H5 OH in 100 cm3 of solution →  M solution
46
.
= 0.0174 M
W13.5
(a)
Volume × Molarity
Number of moles of NaOH = 
1000
20 × 0.1
= 
1000
= 2 × 10-3 mole
= 2 × 10-3 × 40
= 0.08 g
(b)
Volume × Molarity
Number of moles of HNO3 = 
1000
25 × 0.01
= –––––––––
1000
= 2.5 × 10-4 moles
= 2.5 × 10-4 × 63 (Mr HNO3 = 1 + 14 + 3(16) = 63
= 0.0158 g
114
Chemistry Live! – Worked Solutions
(c)
Number of moles of NH3
Volume × Molarity
= 
1000
750 × 0.12
= 
1000
= 0.09 moles
= 0.09 × 17 (Mr NH3 = 14 + 3(1) = 17)
= 1.53 g
(d)
Number of moles of HCl
Volume × Molarity
= 
1000
2000 × 2
= 
1000
= 4 moles
= 4 × 36.5 (Rel. mol mass HCl = 1 + 35.5 = 36.5)
= 146 g
(e)
Number of moles of H2 SO4
Volume × Molarity
= 
1000
20 × 0.1
= 
1000
= 2 × 10-3 moles
= 2 × 10-3 × 98
= 0.196
(Mr H2 SO4 = 2(1) + 32 + 4(16) = 98)
115
Chemistry Live! – Worked Solutions
W13.6
Mg + 2HCl → MgCl2 + H2
Number of moles of HCl
Volume × Molarity
= 
1000
50 × 2
= 
1000
= 0.1 moles
From the balanced equation, 2 moles of hydrochloric acid react with one mole of
magnesium
0.1
⇒  = 0.05 moles of Mg react.
2
= 0.05 × 24
= 1.2 g
W13.7
CaCO3 + 2HCl → CaCl2 + H2 O + CO2
Rel. mol mass CaCO3 = 40 + 12 + 3(16)
= 40 + 12 + 48 = 100
Mass in g
10
Number of moles of CaCO3 =  =  = 0.1
Rel. mol. Mass 100
From the balanced equation, 2 moles of hydrochloric acid react with one mole of CaCO3 .
⇒ (0.1 × 2) moles = 0.2 moles of HCl react.
Volume × Molarity
Number of moles of HCl = 
1000
Number of moles of HCl × 1000
⇒ Volume = 
Molarity
0.2 × 1000
= 
3
= 66.67 cm3
116
Chemistry Live! – Worked Solutions
W13.8
Given:
Vol dil = ?
M dil = 0.01 M
Vol conc = 20 cm3
M conc = 1 M
Vol dil × M dil

1000
Vol dil × 0.01
Vol conc × M conc

1000
=
= 20 × 1
20 × 1
⇒ Vol dil = 
0.01
= 2000 cm3
i.e. 20 cm3 of 1 M acid would have to be diluted to 2000 cm3
W13.9
Given:
Voldil = 2000 cm3
Mdil = 2 M
Volconc = ?
Mconc = 16 M
Vol dil × M dil

1000
2000 × 2
=
Vol conc × M conc

1000
= Vol conc × 16
⇒ Vol conc
2,000 × 2
= 
16
= 250 cm3
i.e. 250 cm3 of 16 M ammonia would have to be diluted to 2 L.
W13.10
Given:
Voldil = 2,500 cm3
Mdil = 20 volume
Volconc = ?
Mconc = 100 volume
Vol dil × M dil

1000
=
Vol conc × M conc

1000
2,500 × 20 = Vol conc × 100
2,500 × 20
⇒ Vol conc = 
100
= 500 cm3
i.e. 500 cm3 of 100 volume H2 O2 would have to be diluted to 2.5 L.
117
Chemistry Live! – Worked Solutions
W13.12
Balanced equation:
Ca(OH)2 + 2HCl → CaCl2 + 2H2 O
Given:
Va = 15.2 cm3
Ma = 0.05 M
na = 2
Vb =25 cm3
Mb = ?
nb = 1
HCl
Ca(OH)2
Va × Ma

na
=
15.2 × 0.05
––––––––––
2
Vb × Mb

nb
25 × Mb
= –––––––
1
15.2 × 0.05
⇒ Mb = ––––––––––
2 × 25
= 0.0152 moles/L
= 0.0152 × 74 (Mr Ca(OH)2 = 74)
= 1.12 g / L
Answers: (a) 0.0152 moles/L
(b) 1.12 g/L
W13.13
(a) Balanced equation:
H2 C2 O4 + 2NaOH → Na2 C2 O4 + 2H2 O
Given:
Va = 18.7 cm3
Ma = ?
na = 2
Vb =25 cm3
Mb = 0.12 M
nb = 1
H2 C2 O4
NaOH
Va × Ma
––––––––
na
Vb × Mb
––––––––
nb
=
18.7 × Ma
–––––––––
1
25 × 0.12
= –––––––––
2
25 × 0.12
⇒ Ma = ––––––––––
2 ×18.7
= 0.08 moles / L
= 0.08 × 90 ( Mr H2 C2 O4 = 90)
= 7.2 g / L
Answers: (a) 0.08 moles/L
(b) 7.2 g/L
118
Chemistry Live! – Worked Solutions
W13.14
Balanced equation:
CH3 COOH + NaOH → CH3 COONa + H2 O
Given:
Va = 20.25 cm3
Ma = ?
na = 1
Vb =25 cm3
Mb = 0.15 M
nb = 1
CH3 COOH
Va × Ma
––––––––
na
NaOH
20.25 × Ma
––––––––––
1
=
Vb × Mb
––––––––
nb
25 × 0.15
= –––––––––
1
25 × 0.15
⇒ Ma = –––––––––
20.25
= 0.1852 moles/L
⇒ Concentration of original solution = 0.1852 × 5 (5 times more concentrated)
= 0.926 moles/L
= 0.926 × 60 (Mr CH3 COOH = 60)
= 55.56 g / L
= 5.56 g /100 cm3
= 5.56 % w / v
Answers = (a) 0.926 moles / L
(b) 55.56 g / L
(c) 5.56 %
119
Chemistry Live! – Worked Solutions
W13.15
Balanced equation:
CH3 COOH + NaOH → CH3 COONa + H2 O
Given:
Va = 12.35 cm3
Ma = ?
na = 1
Vb =25 cm3
Mb = 0.12 M
nb = 1
CH3 COOH
NaOH
Va × Ma
––––––––
na
=
Vb × Mb
–––––––
nb
12.35 × Ma
–––––––––
1
25 × 0.12
= ––––––––
1
25 × 0.12
⇒ Ma = –––––––––
12.35
= 0.2429 moles / L
⇒ Concentration of original solution = 0.2429 × 4 (4 times more concentrated)
= 0.972 moles/L
= 0.972 × 60 (Mr CH3 COOH = 60)
= 58.32 g/L
= 5.83 g/100 cm3
= 5.83 % w/v
Answers = (a) 0.972 moles / L
(b) 58.32 g / L
(c) 5.83 %
120
Chemistry Live! – Worked Solutions
W13.16
(a) Balanced equation:
H2 C2 O4 + 2NaOH → Na2 C2 O4 + 2H2 O + CO2
Given:
Va = 20 cm3
Ma = ?
na = 1
Vb =22.22 cm3
Mb = 0.09 M
nb = 2
H2 C2 O4
Va × Ma
–––––––
na
NaOH
=
20 × Ma
––––––– =
1
Vb × Mb
–––––––
nb
22.22 × 0.09
–––––––––––
2
22.22 × 0.09
⇒ Mb = –––––––––––
2 × 20
= 0.05 moles/L
1 mole H2 C2 O4 = 2(1) + 2(12) + 4(16)
= 2 + 24 + 64 = 90 g
⇒ 0.05 moles H2 C2 O 4 = 0.05 × 90 = 4.5 g
⇒
Mass of water of crystallisation = 6.3 – 4.5
= 1.8 g
1.8
% Water of crystallisation = –––– × 100 = 28.57 %
6.3
(b) Next, calculate the value of x in the formula H2 C2 O4 xH2 O. Note from the above that by
dissolving 6.3 g of the crystals of H2 C2 O4 xH2 O in 1 L of solution, we obtain a solution which
contains 0.05 moles.
Therefore, 0.05 moles H2 C2 O4 .xH2 O = 6.3 g
6.3
1 mole H2 C2 O4 .xH2 O = ––––
0.05
= 126
i.e. Rel. mol mass of H2 C2 O4 .xH2 O = 126
⇒ 2(1) + 2(12) + 4(16) + x(2 +16) = 126
90 + 18x = 126
18x = 36
x= 2
i.e. formula is H2 C2 O4 .2H2 O
Answer:
(a) % Water of crystallisation = 28.57 %
(b) x = 2
121
Chemistry Live! – Worked Solutions
W13.17
Balanced equation:
H2 C2 O4 + 2NaOH → Na2 C2 O4 + 2H2 O + CO2
Given:
Va = 25 cm3
Ma = ?
na = 1
Vb =22.75 cm3
Mb = 0.35 M
nb = 2
H2 C2 O4
NaOH
Va × Ma
–––––––
na
Vb × Mb
–––––––
nb
=
25 × Ma
22.75 × 0.35
–––––––– = –––––––––––
1
2
22.75 × 0.35
⇒ Mb = –––––––––––
2 × 25
= 0.1593 moles/L
0.1593 moles H2 C2 O4 .xH2 O per litre = (0.1593 / 4) moles H2 C2 O4 .xH2 O / 250cm3
= 0.0398 moles H2 C2 O4 .xH2 O/250 cm3
Note from the above that by dissolving 5 g of the crystals of H2 C2 O4 xH2 O in 250 cm3 of
solution, we obtain a solution which contains 0.0398 moles.
Therefore, 0.0398 moles H2 C2 O4 .xH2 O = 5 g
5
1 mole H2 C2 O4 .xH2 O = ––––––
0.0398
= 125.63
i.e. Rel. mol mass of H2 C2 O4 .xH2 O = 125.63
⇒ 2(1) + 2(12) + 4(16) + x(2 +16) = 125.63
90 + 18x = 125.63
18x = 35.63
x = 1.98 ≈ 2
i.e. formula is H2 C2 O4 .2H2 O
Answer: x = 2
Yes – he results obtained by both students are consistent.
122
Chemistry Live! – Worked Solutions
Workbook Chapter 15 – Volumetric Analysis: Oxidation
W15.1
Balanced equation:
MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2 O
Given:Vo =28.5 cm3
Mo = ?
no = 1
Vr =25 cm3
Mred = 0.11 M
nr = 5
MnO4 -
Fe2+
Vo × Mo
Vr × Mred
––––––– = ––––––––
no
nr
28.5 × Mo
–––––––––
1
=
25 × 0.11
––––––––
5
25 × 0.11
⇒ Mo = –––––––––
5 × 28.5
= 0.0193 moles/L
= 0.0193 × 158 (Mr KMnO 4 = 158)
= 3.05 g/L
Answer: (a) 0.0193 moles/L
(b) 3.05 g/L
W15.2
Balanced equation:
MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2 O
Given:
Vo =22.5 cm3
Mo = 0.02 M
no = 1
Vred =25 cm3
Mr = ?
nr = 5
MnO4 -
Fe2+
Vo × Mo
Vr × Mred
––––––– = –––––––––
no
nr
22.5 × 0.02
––––––––––
1
⇒ Mred
=
25 × Mred
–––––––––
5
22.5 × 0.02 × 5
= –––––––––––––
25
= 0.09 moles/L
= 0.09 × 392 (Mr (NH4 )2 (SO4 )FeSO4 .6H2 O
= 392)
= 35.28 g / L
Answer: (a) 0.09 moles/L
(b) 35.28 g/L
123
Chemistry Live! – Worked Solutions
W15.3
(a)
Balanced equation:
MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2 O
MnO4 -
Given:
Vo =23.5 cm3
Mo = 0.025 M
no = 1
Vr =25 cm3
Mred = ?
nr = 5
Vo × Mo
––––––––
no
Fe2+
=
Vr × Mre
–––––––
nr
23.5 × 0.025
25 × Mred
––––––––––– = –––––––––
1
5
⇒ Mred
23.5 × 0.025 × 5
= ––––––––––––––
25
= 0.1175 moles/L
(b)
0.1175 moles FeSO4 / L =
=
=
=
(0.1175 / 4) moles FeSO4 / 250 cm3
0.0294 moles FeSO4 / 250 cm3
0.0294 × 56
(Relative atomic mass Fe = 56)
1.65 g / 250 cm3
1.65
% Fe in steel sample = ––––– ×
100 = 89.19 %
1.85
Answers: (a) 0.1175 moles/L, (b) 1.65 g, 89.19%
124
Chemistry Live! – Worked Solutions
W15.4
(b) Balanced equation:
MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2 O
MnO4 -
Given:
Vo = 19.65 cm3
Mo = 0.024 M
no = 1
Vred = 25 cm3
Mr = ?
nr = 5
Vo × Mo
––––––––
no
Fe2+
19.65 × 0.024
––––––––––––
1
=
⇒ Mred
Vr × Mred
–––––––––
nr
25 × Mred
= ––––––––
5
19.65 × 0.024 × 5
= –––––––––––––––
25
= 0.0943 moles/L
0.0943 moles FeSO4 .xH2 O / L = (0.0943 / 4) moles FeSO4 .xH2 O / 250 cm3
= 0.0236 moles FeSO4 .xH2 O / 250 cm3
i.e. by dissolving 5.71 g of the crystals of FeSO4 .xH2 O in 250 cm3 of solution, we obtain a
solution which contains 0.0236 moles.
Therefore, 0.0236 moles FeSO4 .xH2 O = 5.71 g
1 mole FeSO4 .xH2 O
⇒ 56
5.71
= ––––––
0.0236
= 241.95
+ 32 + 4(16) + x(2 + 16) = 241.95
152 + 18x = 241.95
18x = 89.95
x=5
i.e. formula is FeSO4 .5H2 O
Answer: (b) x = 5
125
Chemistry Live! – Worked Solutions
W15.5
(ii)
392 g of (NH4 )2 (SO4 )FeSO4 .6H2 O in 1 L of solution → 1 M solution
1
1 g of (NH4 )2 (SO4 )FeSO4 .6H2 O in 1 L of solution → ––– M solution
392
1 × 11.76
11.76 g of (NH4 )2 (SO4 )FeSO4 .6H2 O in 1 L of solution → –––––––––– M solution
392
4 × 11.76
10.52 g of (NH4 )2 (SO4 )FeSO4 .6H2 O in 250 cm3 of solution → –––––––––– M solution
392
= 0.12 M
(v)
Balanced equation:
MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2 O
Given:Vo =33.3 cm3
Mo = ?
no = 1
Vred =25 cm3
Mr = 0.12 M
nr = 5
MnO4 Vo × Mo
––––––––
no
Fe2+
=
Vr × Mred
––––––––
nr
33.3 × Mo
–––––––– =
1
25 × 0.12
–––––––––
5
⇒ Mo
25 × 0.12
= –––––––––
5 × 33.3
= 0.018 moles/L
Answer: (ii) 0.12 M, (v) 0.018 M
126
Chemistry Live! – Worked Solutions
W15.6
(vi) (a)
Balanced equation:
MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2 O
Given:Vo = 22.5 cm3
Mo = 0.02 M
no = 1
Vr =25 cm3
Mred = ?
nr = 5
MnO4 -
Fe2+
Vo × Mo
–––––––
no
=
Vr × Mred
–––––––––
nr
22.5 × 0.02
–––––––––––
1
=
25 × Mred
––––––––
5
⇒ Mr
22.5 × 0.02 × 5
= –––––––––––––
25
= 0.09 moles / L
= 0.09 × 284
= 25.56 g / L
Rel. mol mass (NH4 )2 (SO4 )FeSO4 = 2(14) + 8(1) + 32 + 4(16) + 56 + 32 + 4(16) = 284
25.56 g (NH4 )2 (SO4 )FeSO4 / L = (25.56 / 4) g (NH4 )2 (SO4 )FeSO4 / 250 cm3
= 6.39 g (NH4 )2 (SO4 )FeSO4 / 250 cm3
We are told that we dissolved 8.82 g (NH4 )2 (SO4 )FeSO4 . xH2 O / 250 cm3
⇒ mass of H2 O / 250 cm3 = 8.82 – 6.39 = 2.43 g
% water of crystallisation
mass of water
= ––––––––––––––––––––––––––––– × 100
mass of (NH4 )2 (SO4 )FeSO4 . xH2 O
2.43
= ––––– × 100
= 27.55 %
8.82
0.09 moles (NH4 )2 (SO4 )FeSO4 .xH2 O/L = (0.09/4) moles (NH4 )2 (SO4 )FeSO4 .xH2 O / 250 cm3
= 0.0225 moles (NH4 )2 (SO4 )FeSO4 .xH2 O/250 cm3
i.e. by dissolving 8.82 g of the crystals of (NH4 )2 SO4 FeSO4 .xH2 O in 250 cm3 of solution, we
obtain a solution which contains 0.0225 moles.
Therefore, 0.0225 moles (NH4 )2 SO4 FeSO4 .xH2 O = 8.82 g
127
Chemistry Live! – Worked Solutions
8.82
1 mole (NH4 )2 SO4 FeSO4 .xH2 O = ––––––
0.0225
= 392
⇒ 2(14) + 8(1) + 32 + 4(16) + 56 + 32 + 4(16) +x(2 +16)
= 392
284 + 18x = 392
18x = 108
x = 6
i.e. formula is (NH4 )2 SO4 FeSO4 .6H2 O
Answer: (vi) (a) 0.09 M
(b) 25.56 g/L, 27.55%, x = 6.
W15.7
(iii)
(a) Balanced equation:
MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2 O
Given:Vo = 5.47 cm3
Mo = 0.015 M
no = 1
Vr = 20 cm3
Mred = ?
nr = 5
MnO4 Vo × Mo
––––––– =
no
Fe2+
Vr × Mred
––––––––
nr
5.47 × 0.015
20 × Mred
––––––––––– = ––––––––
1
5
5.47 × 0.015 × 5
⇒ Mr = ––––––––––––––
20
= 0.021 moles/L
= 0.021 × 152
= 3.19 g / L
(Rel. mol mass FeSO4 = 56 + 32 + 4(16) = 152)
Therefore, in the 250 cm3 volumetric flask there are 3.19 / 4 g = 0.798 g FeSO4
i.e. 6 tablets contain 0.798 g FeSO4
⇒ 1 tablet contains (0.798 / 6) g = 0.133 g FeSO4
128
Chemistry Live! – Worked Solutions
(b)
Rel. atomic mass Fe
% of Fe in FeSO4 = –––––––––––––––––– × 100
Rel. mol. mass FeSO4
56
= –––– × 100 = 36.84 %
152
⇒ Mass of Fe in each tablet = 36. 84 % of 0.133 g
36.84
= ––––– × 0.133 = 0.049 g
100
(c)
1.47
Mass of 1 tablet = –––– = 0.245 g
6
Mass of FeSO4 in tablet
% FeSO4 in each tablet = –––––––––––––––––––– × 100
Mass of 1 tablet
0.133
= ––––– × 100 = 54.29 %
0.245
Answer: (a) 0.133 g
(b) 0.049 g (c) 54.29%
W15.8
(iii)
(a) Balanced equation:
MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2 O
Given:Vo = 5.75 cm3
Mo = 0.015 M
no = 1
Vr =25 cm3
Mred = ?
nr = 5
MnO4 Vo × Mo
–––––––––
no
Fe2+
=
Vr × Mred
–––––––––
nr
5.75 × 0.015
25 × Mr
––––––––––– = ––––––––
1
5
5.75 × 0.015 × 5
⇒ Mr = ––––––––––––––
25
129
Chemistry Live! – Worked Solutions
= 0.017 moles / L
= 0.017 × 152
= 2.58 g / L
(Rel. mol mass FeSO4 = 56 + 32 + 4(16) = 152)
Therefore, in the 250 cm3 volumetric flask there are 2.58 / 4 = 0.645 g FeSO4
i.e. 5 tablets contain 0.645 g FeSO4
⇒ 1 tablet contains (0.645 / 5) g = 0.129 g FeSO4
(b)
Rel. atomic mass Fe
% of Fe in FeSO4 = –––––––––––––––––– × 100
Rel. mol. mass FeSO4
56
= ––– × 100 = 36.84 %
152
⇒ Mass of Fe in each tablet = 36. 84 % of 0.129 g
36.84
= ––––– × 0.129 = 0.048 g
100
(c)
1.20
Mass of 1 tablet = ––––– = 0.24 g
5
Mass of FeSO4 in tablet
% FeSO4 in each tablet = ––––––––––––––––––––– × 100
Mass of 1 tablet
0.129
= ––––– × 100 = 53.75 %
0.24
Answer: (iii) (a) 0.129 g
130
(b) 0.048 g
(c) 53.75%
Chemistry Live! – Worked Solutions
W15.9
(a) Balanced equation:
I2 + 2S2 O3 2- → S4 O6 2- + 2IGiven:Vo = 25 cm3
Mo = ?
no = 1
Vr =22.15 cm3
Mred = 0.12 M
nr = 2
I2
S2 O3 2-
Vo × Mo
––––––––
no
=
Vr × Mred
––––––––
nr
25 × Mo
––––––––
1
22.15 × 0.12
= ––––––––––––
2
⇒ Mo
22.15 × 0.12
= ––––––––––
2 × 25
= 0.0532 moles/L
= 0.0532 × 254
= 13.51 g/L
Answer: (a) 0.0532 moles/L, (b) 13.51 g/L
131
Chemistry Live! – Worked Solutions
W15.10
(a)
2MnO 4 - + 10I- + 16H+ → 2Mn2+ + 5I2 + 8H2 O
I2 + 2S2 O3 2- → S4 O6 2- + 2IFrom the balanced equations, we see that 2 moles of MnO 4 - produce five moles of I2 and these 5
moles of I2 would then react with 10 moles of S2 O3 2-.
i.e. 2MnO4 - = 5I2 = 10S2 O3 2Given:
Vo = 25 cm3
Mo = 0.022 M
no = 2
Vr = 22.45 cm3
Mr = ?
nr = 10
MnO4 Vo × Mo
–––––––
no
25 × 0.022
––––––––––
2
⇒ Mr
S2 O3 2-
=
Vr × Mr
–––––––
nr
22.45 × Mr
= ––––––––––
10
25 × 0.022 × 10
= ––––––––––––––
2 × 22.45
= 0.1225 moles/L
= 0.1225 × 248
= 30.38 g/L
(Rel. molecular mass Na2 S2 O3 .5H2 O = 248)
Answer: (a) 0.1225 moles/L, (b) 30.38 g/L
132
Chemistry Live! – Worked Solutions
W15.11
(vi) (a)
Balanced equation:
2S2 O3 2- + I2 → S4 O6 2- + 2IGiven:
Vo = 25 cm3
Mo = 0.05 M
no = 1
Vr = 31.2 + 31.3 = 31.25 cm3
2
Mred = ?
nr = 2
S2 O3 2-
I2
Vo × Mo
–––––––
no
V × Mred
= ––––––––
nr
25 × 0.05
31.25 × Mred
–––––––––– = ––––––––––
1
2
⇒ Mred =
25 × 0.05 × 2
––––––––––––
31.25
= 0.08 moles/L
= 0.08 × 248
= 19.84 g / L
(Rel. mol mass Na2 S2 O3 .5H2 O = 248)
Concentration of Na2 S2 O3 .5H2 O in g / L = number of moles /L × rel. mol mass
(vii)
19.84 g Na2 S2 O3 .5H2 O / L = (19.84 / 2) g Na2 S2 O3 .5H2 O / 500 cm3
= 9.92 g Na2 S2 O3 .5H2 O / 500 cm3
9.92
% purity of sample = –––– × 100 = 99.2 %
10
Answer: (vi) (a) 0.08 moles/L, (b) 19.84 g/L
(vii) 99.2%
133
Chemistry Live! – Worked Solutions
W15.12
(a)
ClO - + 2I- + 2H+ → Cl- + I2 + H2 O
2S2 O3 2- + I2 → S4 O6 2- + 2IFrom the balanced equations, we see that 1 mole of ClO - produce 1 moles of I2 and this 1 mole
of I2 would then react with 2 moles of S2 O3 2-.
i.e. 1ClO- = 1I2 = 2S2 O3 2Given:Vo = 25 cm3
Mo = ?
no = 1
Vr =19.56 cm3
Mred = 0.25 M
nr = 2
ClO -
S2 O3 2-
Vo × Mo
––––––– =
no
Vr × Mred
––––––––
nr
25 × Mo
19.56 × 0.25
–––––––– = –––––––––––
1
2
⇒ Mo
⇒ Concentration of original solution
19.56 × 0.25
= –––––––––––
2 × 25
= 0.0978 moles / L
= 0.0978 × 5 (5 times more concentrated)
= 0.489 M
= 0.489 × 74.5 (Mr NaClO = 74.5)
= 36.43 g/L
= 3.643 g/100 cm3
= 3.64 % w / v
Answer: (a) 0.489 moles/L, (b) 36.43 g/L, (c) 3.64%
134
Chemistry Live! – Worked Solutions
Workbook Chapter 16 – Rates of Reaction
W16.3
(ii) graph
(iv) From graph:
1
24 cm of 0.2 M Na2 S2 O3 → –– = 0.01 s-1
t
⇒ t = 100 seconds
W16.4
(iii)
graph
3
(iv)
(a) time required to liberate 8.75 × 10-3 moles of oxygen
Rel. mol. mass O2 = 16 × 2 = 32
Mass of O2 = no. of moles × rel. mol. mass
= 8.75 × 10-3 × 32
= 0.28 g
From graph:
0.28 g → 1.3 minutes
(b) number of moles of oxygen liberated after 2.5 minutes
From graph:
2.5 minutes → 0.39 g
number of moles of O2
mass
= ––––––––––––
Rel. mol. Mass
0.39
= –––– = 1.219 × 10-2
32
(c)
0.435 – 0.2925
0.1425
Instantaneous rate of reaction at 2 minutes = ––––––––––––– = ––––––
3 –1
2
= 0.07125 cm3 / minute
135
Chemistry Live! – Worked Solutions
(v) Total loss in mass = 176.58 – 176.10 = 0.48 g
mass
number of moles of O2 = –––––––––––––
Rel. mol. Mass
0.48
= –––– = 0.015
32
2H2 O2 → 2H2 O + O2
2H2 O2 → O2 (shortened version of equation)
2 moles → 1 mole
⇒ 2 × 0.015 → 0.015 moles
= 0.03 moles H2 O2
W16.5
(ii) graph
(iii) From graph:
1
35 °C → Rate = –– = 0.036 s-1
t
⇒ t = 27.78 s
W16.7
(iii) graph
(iv)
1.75 minutes → 51 cm3 O2
At s.t.p, 1 mole of O2 occupies 22.4 L (22,400 cm3 ).
51
No. of moles of O2 = –––––– = 2.28 × 10-3
22,400
(v)
At s.t.p, 1 mole of O2 occupies 22.4 L (22,400 cm3 ).
⇒ 2 × 10-3 mole of O2 occupies (2 × 10-3 × 22,400) cm3
= 44.8 cm3
3
From graph: 44.8 cm are liberated at a time of 1.2 minutes
(vi)
(a)
Instantaneous rate of reaction at 0.5 minutes
50 – 6.5
= ––––––– = 43.5 cm3 / minute
1–0
= 43.5 cm3 / 60 seconds
= 0.725 cm3 / second
(b)
55 – 55
Instantaneous rate of reaction at 3 minutes = –––––––– = 0 cm3 / minute
3.5 –2.5
= 0 cm3 / 60 seconds
= 0 cm3 / second
136
Chemistry Live! – Worked Solutions
Workbook Chapter 17 – Chemical Equilibrium
W17.7
CH3 COOH + C2 H5OH
Initially:
At Equil. :
Conc. at equil.
º
CH3 COOC2 H5 + H2 O
1
4
0
0
1–x
4–x
x
x
[x / V]
[x / V]
[1 – x / V]
[4 – x / V]
V is the volume of the container which we are not given in the question.
[CH3 COOC2 H5 ] [H2 O]
Kc = ––––––––––––––––––––
[CH3 COOH] [C 2 H5 OH]
(x / V) (x / V)
= –––––––––––––––––– = 4
(1 – x / V) (4 – x / V)
The V terms cancel out in the Kc expression
(x) (x)
∴ ––––––––––– = 4
(1 – x) (4 – x)
x2
–––––––––– = 4
4 – 5x + x2
⇒
⇒
⇒
⇒
⇒
cross multiply
x2 = 4 (4 – 5x + x2 )
x2 = 16 - 20x + 4x2
3x2 – 20x + 16 =0
solve quadratic equation using the formula:
_______
–b ± √b2 - 4ac
x = ––––––––––––– where a = 3, b = –20 and c = 16
2a
______________
20 ± √(-20)2 – 4(3)(16)
x = ––––––––––––––––––––
2(3)
___
20 ± √208
20 ± 14.42
x = ––––––––––––– = ––––––––––
6
6
x = 34.42 = 5.74
or
x = 5.58
6
6
= 0.93
137
Chemistry Live! – Worked Solutions
Chemically, 5.74 does not make sense since we started with only 1 mole of CH3 COOH and 4
moles of C2 H5 OH. Therefore, we take the value of x = 0.93.
Therefore equilibrium concentrations are :CH3 COOH = 1 – x = 1 – 0.93 = 0.07 mol/L
C2 H5 OH = 4 – x = 4 - 0.93 = 3.07 mol/L
CH3 COOC2 H5 = x = 0.93 mol/L
H2 O = x = 0.93 mol/L
Answer: Equilibrium concentration of CH3 COOC2 H5 = 0.93 mol/L
Equilibrium concentration of H2 O = 0.93 mol/L
W17.8
2HI
Initially:
At Equil. :
•
H2
+
I2
0.1
0
0
1 – 0.02
0.01
0.01
[0.01]
[0.01]
Conc. at equil.: [0.08]
Therefore equilibrium concentrations are:
(a) I2 = 0.01 mol/L
(b) HI = 0.08 mol/L
Answer: (a) Equilibrium concentration of I2 = 0.01 mol/L
(b) Equilibrium concentration of HI = 0.08 mol/L
138
= 0.08
Chemistry Live! – Worked Solutions
W17.9
Rel molecular mass CH3 COOH = 12 + 3(1) + 12 + 2(16) + 1
= 12 + 3 + 12 + 32 + 1 = 60
Mass
36
Initial number of moles of CH3 COOH = ––––––––––––––––– = ––– = 0.6
Rel molecular mass
60
Mass
12
At equil. number of moles of CH3 COOH = –––––––––––––––– = ––– = 0.2
Rel molecular mass
60
Rel molecular mass C2 H5 OH = 2(12) + 5(1) + 16 + 1
= 24 + 5 + 16 +1 = 46
Mass
27.6
Initial number of moles of C2 H5 OH = ––––––––––––––––– = ––––– = 0.6
Rel molecular mass
46
CH3 COOH +
Initially:
At Equil. :
C2 H5OH
• CH3COOC2H5 + H2O
0.6
0.6
0
0.6 – 0.4
0.6 – 0.4
0.4
[0.2 / V]
[0.2 / V]
[0.4 / V]
0
0.4 = 0.2
= 0.2
Conc. at equil.
[0.4 / V]
V is the volume of the container which we are not given in the question.
[CH3 COOC2 H5 ] [H2 O]
(0.4 / V) (0.4 / V)
Kc = ––––––––––––––––––– = –––––––––––––––
[CH3 COOH] [C 2 H5 OH]
(0.2 / V) (0.2 / V)
The V terms cancel out in the Kc expression
(0.4) (0.4)
⇒ K c = –––––––– = 4
(0.2) (0.2)
Mass
60
Initial number of moles of CH3 COOH = –––––––––––––––– = ––– = 1.0
Rel molecular mass 60
Rel molecular mass CH3 COOC2 H5 = 12 + 3(1) + 12 + 2(16) + 2(12) + 5(1)
139
Chemistry Live! – Worked Solutions
= 12 + 3 + 12 + 32 + 24 + 5 = 88
Mass
70.4
At equil. number of moles of CH3 COOC2 H5 = ––––––––––––––––– = –––– = 0.8
Rel molecular mass
88
CH3 COOH + C2 H5 OH
• CH3COOC2H5 + H2O
Initially:
1
At Equil. :
C
1 – 0.8
= 0.2
Conc. at equil. [0.2 / V]
0
0
0.8
0.8
[0.8 / V]
[0.8 / V]
C – 0.8
[C – 0.8 / V]
Where C is the initial number of moles of ethanol and V is the volume of the container which
we are not given in the question.
[CH3 COOC2 H5 ] [H2 O]
(0.8 / V) (0.8 / V)
Kc = ––––––––––––––––––– = –––––––––––––––––
[CH3 COOH] [C 2 H5 OH] (0.2 / V) (C – 0.8 / V)
The V terms cancel out in the Kc expression
(0.8) (0.8)
⇒ K c = ––––––––––– = 4
(0.2) (C – 0.8)
0.64
⇒ –––––––––– = 4
0.2C – 0.16
⇒
⇒
⇒
⇒
cross multiply and solve for C
0.64 = 4 (0.2C – 0.16)
0.64 = 0.8C – 0.64
0.8C = 1.28
C = 1.6 moles
Answer Kc = 4
Number of moles of ethanol = 1.6
140
Chemistry Live! – Worked Solutions
W17.10
SO2 (g)
+ NO2 (g) •
SO3(g)
+ NO(g)
[SO3 ] [NO]
Kc = –––––––––––
[SO2 ] [NO2 ]
Rel molecular mass SO2 = 32 + 2(16) = 32 + 32 = 64
Mass
7.68
Initial number of moles of SO2 = ––––––––––––––––– = –––– = 0.12
Rel molecular mass
64
Rel molecular mass NO2 = 14 + 2(16) = 14 + 32 = 46
Mass
4.6
Initial number of moles of NO2 = –––––––––––––––– = ––– = 0.1
Rel molecular mass 46
Rel molecular mass SO3 = 32 + 3(16) = 32 + 48 = 80
Mass
4.8
At equil. number of moles of SO3 = –––––––––––––––– = ––– = 0.06
Rel molecular mass 80
SO2 (g)
Initially:
0.12
At Equil. :
0.12 – 0.06
= 0.06
Conc. at equil.
[0.06 / V]
+
NO2 (g)
0.1
0.1 – 0.06
= 0.04
[0.04 / V]
• SO3(g)
+
0
0
0.06
[0.06 / V]
NO(g)
0.06
[0.06 / V]
V is the volume of the container which we are not given in the question.
[SO3 ] [NO]
(0.06 / V) (0.06 / V)
Kc = ––––––––––– = –––––––––––––––––
[SO2 ] [NO2 ]
(0.06 / V) (0.04 / V)
The V terms cancel out in the Kc expression
(0.06) (0.06)
⇒ K c = ––––––––––– = 1.5
(0.06) (0.04)
Answer Kc = 1.5
141
Chemistry Live! – Worked Solutions
W17.11
2HI
•
H2
+
I2
Initially:
0
2
1
At equil.:
2x
2- x
1- x
Conc. at equil. :
(V =1 L)
[2x]
[H2 ][ I2 ]
Kc = ––––––––
[HI] 2
[2 – x]
[1 –x]
(2 – x) (1 – x)
= ––––––––––– = 0.02
(2x) 2
2 – 3 x + x2
⇒ –––––––––– = 0.02
4x2
cross multiply
2 – 3x + x2 = 0.02 (4x2 )
2 – 3x + x2 = 0.08x2
0.92x2 – 3x + 2 = 0
solve quadratic equation using the formula
_______
– b ± √b2 – 4ac
x = –––––––––––––– where a = 0.92, b = – 3 and c = 2
2a
_______________
3 ± √(– 3)2 – 4(0.92)(2)
⇒
x = –––––––––––––––––––––
2(0.92)
_____
3 ± √1.64
3 ± 1.28
x = –––––––––– = ––––––––
1.84
1.84
⇒
⇒
⇒
⇒
x = 4.28 = 2.33
1.84
or
x = 1.72 = 0.93
1.84
Chemically, 2.33 does not make sense since we started with only 2 moles of H2 . Therefore, we
take the value of x = 0.93.
Therefore equilibrium concentrations are:
HI = 2x = 2(0.93) = 1.86 mol/L
H2 = 2 – x = 2 – 0.93 = 1.07 mol/L
I2 = 1 – x = 1 – 0.93 = 0.07 mol/L
Answer: Equilibrium concentration of HI
= 1.86 mol/L
Equilibrium concentration of H2
= 1.07 mol/L
Equilibrium concentration of I2 = 0.07 mol/Ls
142
Chemistry Live! – Worked Solutions
W17.12
•
Initially:
N2 O4(g)
0.1
2NO2(g)
0
At Equil. :
0.1 - x
2x
Conc. at equil. :
(V = 1 L)
[0.1 - x]
[2x]
[NO2 ] 2
(2x) 2
Kc = ––––––– = –––––––– = 0.36
[N 2 O4 ]
(0.1 - x)
4x2
⇒ –––––– = 0.36
0.1 - x
cross multiply
2
⇒ 4x = 0.36 (0.1 - x)
2
⇒ 4x = 0.036 – 0.36x
2
⇒ 4x + 0.36x – 0.036 = 0
solve quadratic equation using the formula
_______
-b ± √b2 - 4ac
x = –––––––––––– where a = 4, b = 0.36 and c = -0.036
2a
__________________
-0.36 ± √(0.36)2 – 4(4)( -0.036)
⇒
x = ––––––––––––––––––––––––––
2(4)
____
-0.36 ± √0.71
-0.36 ± 0.84
x = ––––––––––––– = ––––––––––––
8
8
⇒
x = 0.48 = 0.06
8
or
x = -1.2
8
= -0.15
Chemically, -0.15 does not make sense, therefore, we take the value of x = 0.06.
Therefore equilibrium concentrations are:N2 O4 = 0.1 – x = 0.1 – 0.06 = 0.04 mol/L
NO2 = 2x = 2 (0.06) = 0.12 mol/L
Answer Equilibrium concentration of N2 O4 = 0.04 mol/L
Equilibrium concentration of NO2 = 0.12 mol/L
143
Chemistry Live! – Worked Solutions
Workbook Chapter 18 – pH and Indicators
W 18.2
(a) pH = - log 10 [H+] = - log 10 (10-5 )
=5
+
(b) pH = - log 10 [H ] = - log 10 (8.33 × 10-3 )
= 2.08
+
(c) pH = - log 10 [H ] = - log 10 (2.25 × 10-1 )
= 0.65
+
(d) pH = - log 10 [H ] = - log 10 (4.4 × 10-12 )
= 11.36
+
(e) pH = - log 10 [H ] = - log 10 (3.9 × 10-9 )
= 8.41
W18.3
(a) pH = - log 10 [H+] = 1.16
log 10 [H+] = - 1.16
+
⇒ [H ] = antilog (- 1.16)
= 0.07 mol/L
(b) pH = - log 10 [H+] = 6.51
log 10 [H+] = - 6.51
+
⇒ [H ] = antilog (- 6.51)
= 3.09 × 10-7 mol/L
(c) pH = - log 10 [H+] = 14
log 10 [H+] = - 14
+
⇒ [H ] = antilog (- 14)
= 1 × 10-14 mol/L
(d) pH = - log 10 [H+] = 0.21
log 10 [H+] = - 0.21
+
⇒ [H ] = antilog (- 0.21)
= 0.62 mol/L
(e) pH = - log 10 [H+] = 10.48
log 10 [H+] = - 10.48
+
⇒ [H ] = antilog (- 10.48)
= 3.31 × 10-11 mol/L
144
Chemistry Live! – Worked Solutions
W18.4
(a) H2 SO4 → 2H+ + SO4 21 mole → 2 moles
⇒ 0.1 mole → 0.1 × 2 mole
= 0.2 mole
pH = - log 10 [H+] = -log 10 (0.2) = 0.70
(b) HCl → H+ + Cl1 mole → 1 mole
⇒ 0.2 mole → 0.2 mole
pH = - log 10 [H+] = -log 10 (0.2) = 0.70
(c) HCl → H+ + Cl1 mole → 1 mole
⇒ 0.05 mole → 0.05 mole
pH = - log 10 [H+] = -log 10 (0.05) = 1.30
(d) 0.007 g KOH / 500 cm3 = (0.007 × 2) g KOH / L = 0.014 g KOH / L
Rel molecular mass KOH = 39 + 16 + 1 = 56
Mass in 1 L
0.014
Number of moles of KOH / L = –––––––––––––––– = –––––– = 0.00025
Rel molecular mass
56
KOH → K+ + OH1 mole → 1 mole
⇒ 0.00025 mole → 0.00025 mole
pOH = - log 10 [OH-] = -log 10 (0.00025) = 3.60
pH = 14 – pOH =14 –3.60 = 10.40
(e) 0.049 g H2 SO4 / 200 cm3 = (0.049 × 5) g H2 SO4 / L = 0.245 g H2 SO4 / L
Rel molecular mass H2 SO4 = 2(1) + 32 + 4(16) = 98
Mass in 1 L
0.245
Number of moles of H2 SO4 / L = –––––––––––––––– = ––––– = 0.0025
Rel molecular mass
98
H2 SO4 → 2H+ + SO4 21 mole → 2 mole
⇒ 0.0025 mole → 0.0025 × 2 mole
= 0.005 mole
+
pH = - log 10 [H ] = -log 10 (0.005) = 2.30
145
Chemistry Live! – Worked Solutions
W18.5
(a) NaOH → Na+ + OH1 mole → 1 mole
⇒ 0.33 mole → 0.33 mole
pOH = - log 10 [OH-] = -log 10 (0.33) = 0.48
pH = 14 – pOH = 14 – 0.48 = 13.52
(b) KOH → K+ + OH1 mole → 1 mole
⇒ 0.001 mole → 0.001 mole
pOH = - log 10 [OH-] = -log 10 (0.001) = 3
pH = 14 – pOH = 14 – 3 = 11
(c) 2 g NaOH / 250 cm3 = (2 × 4) g NaOH / L = 8 g NaOH / L
Rel molecular mass NaOH = 23 + 16 + 1 = 40
Mass in 1L
8
Number of moles of NaOH / L = –––––––––––––––– = ––– = 0.2
Rel molecular mass 40
NaOH → Na+ + OH1 mole → 1 mole
⇒ 0.2 mole → 0.2 mole
pOH = - log 10 [OH-] = -log 10 (0.2) = 0.70
pH = 14 – pOH = 14 – 0.70 = 13.30
(d) KOH → K+ + OH1 mole → 1 mole
-5
-5
⇒ 5 ×10 mole → 5 ×10 mole
pOH = - log 10 [OH-] = -log 10 (5 ×10-5 ) = 4.30
pH = 14 – pOH =14 – 4.30 = 9.70
(e) 0.008 g KOH / 200 cm3 = (0.008 × 5) g KOH / L = 0.04 g KOH / L
Rel molecular mass KOH = 39 + 16 + 1 = 56
Mass in 1 L
0.04
Number of moles of KOH / L = –––––––––––––––– = ––––– = 7.14 ×10-4
Rel molecular mass
56
KOH → K+ + OH1 mole → 1 mole
-4
-4
⇒ 7.14 ×10 mole → 7.14 ×10 mole
pOH = - log 10 [OH ] = -log 10 (7.14 ×10-4 ) = 3.15
pH = 14 – pOH =14 –3.15 = 10.85
146
Chemistry Live! – Worked Solutions
W18.6
Hydroxide ion concentration can be found by substituting into the formula:
__________
[OH ] = √ Kb × Mbase
_______________
= √ 1.8 ×10-5 × 0.05
_________
= √ 9 ×10-7
= 9.49 ×10-4 mol/L
pOH = - log 10 [OH-] = -log 10 (9.49 ×10-4 ) = 3.02
pH = 14 – pOH =14 – 3.02 = 10.98
Answer: pH = 10.98
W18.7
Hydrogen ion concentration can be found by substituting into the formula:__________
+
[H ] = √ Ka × Macid
____________
= √1.8 × 10-4 × 0.1
_________
= √ 1.8 × 10-5
= 4.24 × 10-3 mol/L
pH = - log 10 [H+] = -log 10 (4.24 × 10-3 ) = 2.37
Answer: pH =2.37
W 18.8
Kw = [H+] [OH-]
[H+] = [OH-]
+ 2
⇒ K w = [H ]
At 25 o C
Kw = 10-14 = [H+]2
_____
⇒
[H ] = √10-14 = 1×10-7 mol/L
+
pH = - log 10 [H+] = -log 10 (1×10-7 ) = 7
At 57 o C
Kw = 9 ×10-14 = [H+]2
________
⇒
[H+] = √9 ×10-14 = 3 ×10-7 mol/L
pH = - log 10 [H+] = -log 10 (3×10-7 ) = 6.52
Answer: pH at 25 o C = 7
pH at 57 o C = 6.52
147
Chemistry Live! – Worked Solutions
W18.9
8 g CH3 COOH / 250 cm3 = (8 × 4) g CH3 COOH / L = 32 g CH3 COOH / L
Rel molecular mass CH3 COOH = 12 + 3(1) + 12 + 2(16) + 1 = 60
Mass in 1 L
32
Number of moles of CH3 COOH / L = ––––––––––––––––– = ––– = 0.533
Rel molecular mass
60
CH3 COOH → CH3 COO-
+
H+
Hydrogen ion concentration can be found by substituting into the formula:
_________
+
[H ] = √ Ka × Macid
_______________
= √1.8 × 10-5 × 0.533
___________
= √ 9.594 × 10-6
= 3.1 × 10-3 mol/L
pH = - log 10 [H+] = -log 10 (3.1 × 10-3 ) = 2.51
Answer: pH = 2.51
W18.10
3 g CH3 COOH / 250 cm3 = (3 × 4) g CH3 COOH / L = 12 g CH3 COOH / L
Rel molecular mass CH3 COOH = 12 + 3(1) + 12 + 2(16) + 1 = 60
Mass in 1 L
12
Number of moles of CH3 COOH / L = ––––––––––––––––– = ––– = 0.2
Rel molecular mass
60
Hydrogen ion concentration can be found by substituting into the formula:
_________
+
[H ] = √ Ka × Macid
_____________
= √ 1.8 × 10-5 × 0.2
_________
= √ 3.6 × 10-6
= 1.90 × 10-3 mol/L
pH = - log 10 [H+] = -log 10 (1.9 × 10-3 )
= 2.72
Answer: pH = 2.72
148
Chemistry Live! – Worked Solutions
W18.11
Acid concentration can be found by substituting into the formula:
__________
+
[H ] = √ Ka × Macid
Square both sides
[H+] 2 = Ka × Macid
(1.5 × 10-4 )2 = 1.8 × 10-5 × Macid
(1.5 × 10-4 )2 2.25 × 10-8
-3
⇒ Macid = –––––––––– = ––––––––– = 1.25 ×10 mol/L
1.8 ×10-5
1.8 ×10-5
pH = - log 10 [H+] = -log 10 (1.5 × 10-4 ) = 3.82
Answer: Macid = 1.25 ×10-3 mol/L
pH = 3.82
W 18.12
5.5 g H3 BO3 / L
Rel molecular mass H3 BO3 = 3(1) + 11 + 3(16) = 62
Mass in 1L
5.5
Number of moles of HNO3 / L = –––––––––––––––– = –––– = 0.089
Rel molecular mass
62
H3 BO3 → H+ + H2 BO3 1 mole → 1 mole
⇒ 0.089 mole → 0.089 mole
pH = - log 10 [H+] = - log 10 (0.089) = 1.05
Answer: pH = 1.05
W18.13
0.1 M HX acid is 3.5 % dissociated i.e. 3.5 % of 0.1 mole
3.5
–––– × 0.1 = 3.5 × 10-3 mole of H+ ions are formed
100
Ka can be found by substituting into the formula:
__________
[H+] = √ Ka × Macid
Square both sides
[H+] 2 = Ka × Macid
(3.5 × 10-3 )2 = Ka × 0.1
-4
⇒ K a = 1.225 × 10
Answer: Ka = 1.225 × 10-4
149
Chemistry Live! – Worked Solutions
W 18.14
(ii) Hydrogen ion concentration can be found by substituting into the formula:
_________
+
[H ] = √ Ka × Macid
____________
= √ 8 ×10-3 × 0.01
_________
= √ 8 × 10-5
= 8.94 ×10-3 mol/L
pH = - log 10 [H+] = -log 10 (8.94 ×10-3 ) = 2.05
Answer: pH = 2.05
W 18.17
(ii) Hydrogen ion concentration can be found by substituting into the formula:
_________
[H+] = √Ka × Macid
_____________
= √2 ×10-5 × 0.01
_______
= √2 ×10-7
= 4.47 ×10-4 mol/L
pH = - log 10 [H+] = -log 10 (4.47 ×10-4 ) = 3.35
Answer: pH =3.35
W 18.20
pH = - log 10 [H+] = 2.7
log 10 [H+] = - 2.7
+
⇒ [H ] = antilog (- 2.7)
= 2 × 10-3 mol/L
HA → H+ + A1 mole → 1 mole
-3
-3
⇒ 2 × 10 mole → 2 × 10 mole
⇒
[HA] = 2 × 10-3 mol/L
Ka of HX can be found by substituting into the formula:
__________
+
[H ] = √ Ka × Macid
Square both sides
[H+] 2 = Ka × Macid
(2 × 10-3 )2 = Ka × 0.5
-6
⇒ K a = 8 × 10
Answer: [HA] = 2 × 10-3 mol/L
Ka = 8 × 10-6
150
Chemistry Live! – Worked Solutions
Workbook Chapter 19 – Environmental Chemistry – Water
W19.2
(e)
Given:VCa = 50 cm3
MCa = ?
nCa = 1
Ved = 18 cm3
Med = 0.01 M
ned = 1
Ca2+
edta
VCa × MCa
––––––––––
nCa
Ved × Med
–––––––––
ned
=
50 × MCa
18 × 0.01
––––––––– = ––––––––
1
1
18 × 0.01
⇒ MCa = –––––––––
50
= 3.6 × 10-3 moles/L CaCO3
= 3.6 × 10-3 × 100 g/L CaCO3 (Mr CaCO3 = 100)
= 0.36 g/L CaCO3
= 0.36 × 1000 mg/L CaCO3
= 360 mg/L CaCO3
= 360 p.p.m. CaCO3
Answer Total hardness of water = 360 p.p.m.
W19.3
(c)
(i) Total hardness
Given:VCa = 50 cm3
MCa = ?
nCa = 1
Ved = 11.5 cm3
Med = 0.01 M
ned = 1
Ca2+
VCa × MCa
–––––––––
nCa
edta
Ved × Med
= ––––––––––
ned
50 × MCa
11.5 × 0.01
––––––––– = ––––––––––
1
1
11.5 × 0.01
⇒ MCa = ––––––––––
50
= 2.3 × 10-3 moles/L CaCO3
= 2.3 × 10-3 × 100 g/L CaCO3 (Mr CaCO3 = 100)
= 0.23 g/L CaCO3
= 0.23 × 1000 mg/L CaCO3
= 230 mg/L CaCO3
= 230 p.p.m. CaCO3
151
Chemistry Live! – Worked Solutions
(ii) Permanent Hardness, i.e. hardness which is not removed on boiling
Given:Ca2+
edta
VCa = 50 cm3
MCa = ?
VCa × MCa
Ved × Med
nCa = 1
––––––––– = ––––––––––
Ved = 6.5 cm3
nCa
ned
Med = 0.01 M
ned = 1
50 × MCa
6.5 × 0.01
–––––––– = –––––––––
1
1
6.5 × 0.01
⇒ MCa = –––––––––
50
= 1.3 × 10-3 moles/L CaCO3
= 1.3 × 10-3 × 100 CaCO3 (Mr CaCO3 = 100)
= 0.13 g/L CaCO3
= 0.13 × 1000 mg/L CaCO3
= 130 mg/L CaCO3
= 130 p.p.m. CaCO3
(iii) Temporary hardness, i.e. hardness which is removed on boiling.
Temporary hardness = Total hardness – Permanent hardness
= 230 –130 = 100 p.p.m.
Answer: (i) Total hardness of water = 230 p.p.m.
(ii) Permanent hardness of water = 130 p.p.m.
(iii) Temporary hardness of water = 100 p.p.m.
152
Chemistry Live! – Worked Solutions
W19.4
(iv)
Given:
VCa = 50 cm3
MCa = ?
nCa = 1
Ved = 15 cm3
Med = 0.01 M
ned = 1
Ca2+
edta
VCa × MCa
–––––––––
nCa
Ved × Med
–––––––––
ned
=
50 × MCa
––––––––– =
1
15 × 0.01
–––––––––
1
15 × 0.01
⇒ MCa = ––––––––
50
= 3 × 10-3 moles/L CaCO3
= 3 × 10-3 × 100 g/L CaCO3 (Mr CaCO3 = 100)
= 0. 3 g/L CaCO3
= 0.3 × 1000 mg/L CaCO3
= 300 mg/L CaCO3
= 300 p.p.m. CaCO3
Answer: Total hardness of water = 300 p.p.m.
W19.5
(iv)
Given:VCa = 50 cm3
MCa = ?
nCa = 1
Ved = 9.5 cm3
Med = 0.01 M
ned = 1
Ca2+
edta
VCa × MCa
–––––––––
nCa
Ved × Med
–––––––––
ned
50 × MCa
–––––––– =
1
=
9.5 × 0.01
––––––––––
1
9.5 × 0.01
⇒ MCa = –––––––––
50
= 1.9 × 10-3 moles/L
= 1.9 × 10-3 × 40 g/L Ca2+
= 0.076 g/L Ca2+ …………. (a)
= 1.9 × 10-3 × 100 g/L CaCO3
= 0.19 g/L CaCO3
= 0.19 × 1000 mg/L
= 190 mg/L
= 190 p.p.m. CaCO3 ………(b)
Answer (a) Total hardness of water = 0.076 g of Ca2+/L
(b) Total hardness of water = 190 p.p.m. CaCO3
153
Chemistry Live! – Worked Solutions
W19.6
Total suspended solids = 1.47 g / 750 cm3
4
= 1.47 × ––– g/L
3
4
= 1.47 × ––– × 1,000 mg/L
3
= 1960 p.p.m.
Total dissolved solids = 0.73 g / 200 cm3
= 0.73 × 5 g/L
= 0.73 × 5 × 1,000 mg/L
= 3650 p.p.m.
Answer: Total suspended solids = 1960 p.p.m.
Total dissolved solids = 3650 p.p.m.
W19.8
Given:Vo = 200 cm3
Mo = ?
no = 1
Vr = 22.7 cm3
Mr = 0.01 M
nr = 4
S2 O3 2-
O2
Vo × Mo
––––––––
no
=
Vr × Mr
––––––––
nr
200 × Mo
–––––––– =
1
22.7 × 0.01
––––––––––
4
22.7 × 0.01
⇒ Mo = ––––––––––––
4 × 200
=
=
=
=
=
=
154
2.8375 × 10-4 moles/L
2.8375 × 10-4 × 32 g/L (Rel molecular mass O2 =32)
9.08 × 10-3 g/L
9.08 × 10-3 × 1000 mg/L
9.08 mg/L
9.08 p.p.m.
Chemistry Live! – Worked Solutions
After 5 days
Given:
Vo = 200 cm3
Mo = ?
no = 1
Vr = 4.8 cm3
Mr = 0.01 M
nr = 4
S2 O3 2-
O2
Vo × Mo
––––––––– =
no
Vr × Mr
––––––––
nr
200 × Mo
–––––––– =
1
4.8 × 0.01
––––––––––
4
4.8 × 0.01
⇒ Mo = ––––––––
4 × 200
= 6 × 10-5 moles/L
= 6 × 10-5 × 32 g/L (Rel molecular mass O2 =32)
= 1.92 × 10-3 g/L
= 1.92 × 10-3 × 1000 mg/L
= 1.92 mg/L
= 1.92 p.p.m.
B.O.D. in diluted sample = 9.08 –1.92 = 7.16
But since the original water sample was diluted ten times
⇒ B.O.D. of original water sample = 7.16 × 10 = 71.6 p.p.m.
Answer: B.O.D. = 71.6 p.p.m.
155
Chemistry Live! – Worked Solutions
Workbook Chapter 21 – Fuels and Heats of Reaction
W21.12
Mass of solution = 100 g = 0.1 kg
Temperature rise = 21.1 – 14.3 = 6.8 °C
Specific heat capacity = 4,060 J kg-1 K-1
Heat liberated = mass × specific heat capacity × temp. rise
= 0.1 × 4,060 × 6.8
= 2,760.8 J
volume × molarity 50 × 1
Number of moles of HCl neutralised = –––––––––––––––– = –––––– = 0.05
1000
1000
i.e. 0.05 mole HCl neutralised liberates 2,760.8 J
2,760.8
1 mole HCl neutralised liberates ––––––– = 55,216 J
0.05
Since for HCl, 1 mole of acid gives 1 mole of H+ ions
∴Heat of neutralisation = - 55.216 k J mol-1
i.e. HCl + NaOH → NaCl + H2 O
∆H = - 55.216 k J mol-1
Answer: Heat of neutralisation = - 55.216 k J mol-1
W21.13
Mass of solution = 100 g = 0.1 kg
Temperature rise = 13 °C
Specific heat capacity = 4,200 J kg-1 K-1
Heat liberated = mass × specific heat capacity × temp. rise
= 0.1 × 4,200 × 13
= 5,460 J
volume × molarity 50 × 2
Number of moles of HNO3 neutralised = –––––––––––––––– = –––––– = 0.1
1000
1000
i.e. 0.1 mole HNO3 neutralised liberates 5,460 J
⇒
5,460
1 mole HNO3 neutralised liberates –––––– = 54,600 J
0.1
Since for HNO3 , 1 mole of acid gives 1 mole of H+ ions
∴Heat of neutralisation = - 54.6 k J mol-1
HNO3 + KOH → KNO3 + H2 O
Answer: Heat of neutralisation = - 54.6 k J mol-1
156
Chemistry Live! – Worked Solutions
W21.14
Required Equation:- CH4 (g) + 4Cl2 (g) → CCl4 (l) + 4HCl (g)
(a) reversed
(b)
(c) × 4
CH4 (g)
→ C(s) + 2H2 (g)
C(s) + 2Cl2 (g) → CCl4 (l)
2H2 (g) + 2Cl2 (g) → 4HCl (g)
∆H =?
∆H = 74.9 kJ mol-1
∆H = -139 kJ mol-1
∆H = -369.2 kJ mol-1
CH4 (g) + C(s) + 2Cl2 (g) + 2H2 (g) + 2Cl2 (g) → C(s) + 2H2 (g) + CCl4 (l) + 4HCl (g)
i.e. CH4 (g) + 4Cl2 (g) → CCl4 (l) + 4HCl (g)
∆H = -433.3 kJ mol-1
Answer: ∆H = -433.3 kJ mol-1
W21.15
Required Equation: CH3 OH (l) + 1½O
(c) reversed
(b) × 2
(a)
2 (g)
→ CO2 (g) + 2H2 O(l)
CH3 OH (l) → C(s) + 2H2 (g) + O
½
2H2 (g) + O2 (g) → 2H2 O(l)
C(s) + O2 (g) → CO2 (g)
∆H = 250 kJ mol-1
∆H = -572 kJ mol-1
∆H = -394 kJ mol-1
2 (g)
CH3 OH (l) + 2H2 (g) + O2 (g) + C(s) + O2 (g) → C(s) + 2H2 (g) + O
½
i.e. CH3 OH (l) + 1½O
2 (g)
→ CO2 (g) + 2H2 O(l)
∆H =?
2 (g)
+ 2H2 O(l) + CO2 (g)
∆H = -716 kJ mol-1
Answer: ∆H = -716 kJ mol-1
W21.16
Required Equation:- 2CH4 (g) + O2 (g) →2CH3 OH(l)
(a) × 2
(c)
(b) × 2
∆H =?
2CH4 (g) + 2H2 O (g) → 2CO (g) + 6H2 (g) ∆H = 412 kJ mol-1
2H2 (g) + O2 (g)
→ 2H2 O (l)
∆H = -434 kJ mol-1
4H2 (g) + 2CO (g)
→ 2CH3 OH (g)
∆H = -256 kJ mol-1
2CH4 (g) + 2H2 O (g) + 2H2 (g) + O 2 (g) + 4H2 (g) + 2CO (g) → 2CO (g) + 6H2 (g) +2H2 O (l) + 2CH3 OH (g)
i.e
2CH4 (g) + O2 (g) → 2CH3 OH(l)
∆H = -278 kJ mol-1
Answer: ∆H = -278 kJ mol-1
W21.17
Required Equation: C (s) + 2H2 (g) → CH4 (g)
∆H =?
(b) × 2
(c)
(a) reversed
∆H = -571.6 kJ mol-1
∆H = -393.5 kJ mol-1
∆H = 890.4 kJ mol-1
2H2 (g) + O2 (g) → 2H2 O (l)
C (s) + O2 (g)
→ CO2 (g)
CO2 (g) + 2H2 O (l) → CH4 (g) + 2O2 (g)
2H2 (g) + O2 (g) + C (s) + O2 (g) + CO2 (g) + 2H2 O (l) → 2H2 O (l) + CO2 (g) + CH4 (g) + 2O2 (g)
i.e. C (s) + 2H2 (g) → CH4 (g)
∆H = -74.7 kJ mol-1
Answer: ∆H = -74.7 kJ mol-1
157
Chemistry Live! – Worked Solutions
W21.18
Required Equation:- 3C (s) + 4H2 (g) → C3 H8 (g)
(b) × 3
(c) × 4
(a) reversed
∆H =?
3C (s) + 3O2 (g)
→ 3CO2 (g)
∆H = -1,182 kJ mol-1
4H2 (g) + 2O2 (g) → 4H2 O (l)
∆H = -1,144 kJ mol-1
3CO2 (g) + 4H2 O (l) → C3 H8 (g) + 5O2 (g) ∆H = 2,220 kJ mol-1
3C (s) + 3O2 (g) + 4H2 (g) + 2O2 (g) +3CO2 (g) + 4H2 O (l) → 3CO2 (g) + 4H2 O (l) + C 3 H8 (g) + 5O2 (g)
i.e. 3C (s) + 4H2 (g) → C3 H8 (g)
∆H = -106 kJ mol-1
Answer: ∆H = -106 kJ mol-1
W21.19
Required Equation:- 2C (s) + H2 (g) → C2 H2 (g)
∆H =?
(b) × 2
2C (s) + 2O2 (g)
→ 2CO2 (g)
∆H = -788 kJ mol-1
(c)
H2 (g) + ½O2 (g) → H2 O (l)
∆H = -286 kJ mol-1
(a) reversed 2CO2 (g) + H2 O (l) → C2 H2 (g) + 2½O 2 (g) ∆H = 1,299 kJ mol-1
2C (s) + 2O2 (g) + H2 (g) + O
½
i.e.
2 (g)
+ 2CO2 (g) + H2 O (l) → 2CO2 (g) + H2 O (l) + C2 H2 (g) + 2½O
2C (s) + H2 (g) → C2 H2 (g)
∆H = 225 kJ mol
-1
Answer: ∆H = 225 kJ mol-1
W21.20
Required Equation:- N
½
(a) reversed × ¼ N
½
(b) × 1½
1½H
N
½
2 (g)
i.e.
2 (g)
2 (g)
2 (g)
→ NH3 (g)
∆H =?
+ 1½H 2 O (l) → NH3 (g) + ¾O
O 2 (g) → 1½H 2 O (l)
2 (g) + ¾
2 (g)
+ 1½H 2 O (l) +1½H
N
½
+ 1½H
+ 1½H
2 (g)
2 (g)
+ ¾O
2 (g)
2 (g)
→ NH3 (g) + ¾O
→ NH3 (g)
∆H = 317 kJ mol-1
∆H = -428.7 kJ mol-1
2 (g)
+1½H 2 O (l)
∆H = -111.7 kJ mol-1
Answer: ∆H = -111.7 kJ mol-1
W21.21
Required Equation:- Ca (s) +C (s) + 1½O
(a) × ½
(b)
2 (g)
∆H =?
∆H = -320 kJ mol-1
∆H = -393 kJ mol-1
+ CO2 + C (s) + O 2 (g) → CaCO3 (s) + CO2 (g)
i.e. Ca (s) +C (s) + 1½O
2 (g)
→ CaCO3 (s)
Answer: ∆H = -713 kJ mol-1
158
→ CaCO3 (s)
Ca (s) + O
½ 2 (g) + CO2 → CaCO3 (s)
C (s) + O2 (g) → CO2 (g)
_________________
Ca (s) + O
½
2 (g)
∆H = -713 kJ mol-1
2 (g
Chemistry Live! – Worked Solutions
W21.22
Required Equation:- C2 H2 (g) + HCN (g) → C3 H3 N (g)
∆H =?
Given
(a)
2C (s) + H2 (g)
→ C2 H2 (g)
∆H = 233 kJ mol-1
(b)
H
½ 2 (g) + C (s) + N
½ 2 (g) → HCN (g)
∆H = 114.5 kJ mol-1
(c)
3C (s) + 1½H 2 (g) + N
½ 2 (g) → C3 H3 N (g)
∆H = 194.5 kJ mol-1
(a) reversed
(b) reversed
(c)
C2 H2 (g) → 2C (s) + H2 (g)
∆H = -233 kJ molHCN (g→ H
½ 2 (g) + C (s) + N
½ 2 (g) ∆H = -114.5 kJ mol-1
3C (s) + 1½H 2 (g) + N
½ 2 (g) → C3 H3 N (g) ∆H = 194.5 kJ mol-1
C2 H2 (g) +HCN (g) + 3C (s) + 1½H
i.e
2 (g)
+N
½
2 (g)
→ 2C (s) + H2 (g) + H
½
C2 H2 (g) + HCN (g) → C3 H3 N (g)
2 (g)
+ C (s) + N
½
2 (g) +C3 H3 N (g)
∆H = -153 kJ mol-1
Answer: ∆H = -153 kJ mol-1
W21.23
Required Equation:- H2 (g) + S (s) + 2O2 (g) → H2 SO4 (l)
∆H =?
Given
(a)
H2 (g) + O
½ 2 (g) → H2 O (l)
∆H = -286 kJ mol-1
(b)
S (s) + O2 (g)
→ SO2 (g)
∆H = -297 kJ mol-1
(c)
2SO2 (g) + O 2 (g) → 2SO3 (g)
∆H = -196 kJ mol-1
(d)
SO3 (g) + H2 O (l) → H2 SO4 (l)
∆H = -133 kJ mol-1
(a)
(b)
(c) × ½
(d)
H2 (g) + O
½ 2 (g) → H2 O (l)
S (s) + O2 (g)
→ SO2 (g)
SO2 (g) + O
½ 2 (g) → SO3 (g)
SO3 (g) + H2 O (l) → H2 SO4 (l)
∆H = -286 kJ mol-1
∆H = -297 kJ mol-1
∆H = -98 kJ mol-1
∆H = -133 kJ mol-1
H2 (g) + O
½ 2 (g) + S (s) + O2 (g) + SO2 (g) + O
½ 2 (g) + SO3 (g) + H2 O (l) → H2 O (l) + SO2 (g) + SO3 (g)
+H2 SO4 (l)
i.e.
H2 (g) + S (s) + 2O2 (g) → H2 SO4 (l) ∆H = -814 kJ mol-1
Answer: ∆H = -814 kJ mol-1
159
Chemistry Live! – Worked Solutions
Chapter 24 –Stoichiometry II
W24.1
Relative molecular mass FeS2 = 56 + 2(32) = 120
Relative molecular mass SO2 = 32 + 2(16) = 64
Iron pyrites is the limiting reagent as oxygen is in excess.
Mass
5000
Number of moles of iron pyrites = –––––––––––––––– = –––––– = 416.67
Rel molecular mass
12
4FeS2 + 11O2 → 2Fe2 O3
+ 8SO2
4 moles → 8 moles
⇒ 416.67 moles → 416.67 × 2 moles
= 833.34 moles
= 833.34 × 64
= 53,333.76 g
i.e. the theoretical yield of SO2 is 53,333.76 g
Actual yield of product
Percentage yield = ––––––––––––––––––––––––
Theoretical yield of product
×
100
2725
= ––––––––– × 100
53,333.76
= 5.11%
Answer : Percentage yield of sulphur dioxide is 5.11%.
W24.2
Relative molecular mass NH4 Cl = 14 + 4(1) + 35.5 = 53.5
Relative molecular mass Ca(OH)2 = 40 + 2(17) = 74
Relative molecular mass NH3 = 14 + 3(1) = 17
Mass
20
Number of moles of ammonium chloride = ––––––––––––––––– = ––––– = 0.37
Rel molecular mass
53.5
Mass
20
Number of moles of calcium hydroxide = ––––––––––––––––– = –––– = 0.27
Rel molecular mass
74
2NH4 Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2 O
2 moles
1 mole
⇒ 0.37 moles
0.37 moles
2
= 0.185 moles
160
Chemistry Live! – Worked Solutions
0.185 moles of calcium hydroxide would react with 0.37 moles of ammonium chloride.
However, there is 0.27 moles of calcium hydroxide present, i.e. the calcium hydroxide is present
in excess.
Therefore the ammonium chloride is the limiting reactant, so we calculate the percentage yield
of ammonia on the amount of ammonium chloride present.
2NH4 Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2 O
2 moles → 2 moles
⇒ 0.37 moles → 0.37 moles
= 0.37 × 17
= 6.29 g
i.e. the theoretical yield of ammonia is 6.29 g
⇒ Maximum mass of ammonia produced = 6 g
(to nearest gram)
Answer: The maximum mass of ammonia produced in the reaction is 6 g.
W24.3
Assume ethanol is the limiting reagent.
Relative molecular mass C2 H5 OH = 2(12) + 5(1) + 16 + 1 = 46
Relative molecular mass CH3 COOH = 12 + 3(1) +12 +2(16) +1 = 60
Mass
24
Number of moles of ethanol = –––––––––––––––– = –––– = 0.52
Rel molecular mass
46
3C2 H5 OH + 2Cr2 O7 2- + 16H+ → 4Cr3+ + 3CH3 COOH + 11H2 O
3 moles → 3 moles
⇒ 0.52 moles → 0.52 moles
= 0.52 × 60
= 31.2 g
i.e. the theoretical yield of ethanoic acid is 31.2 g
Percentage yield =
Actual mass yield of product
––––––––––––––––––––––––– ×
Theoretical yield of product
100
28.5
= ––––– × 100
31.2
= 91.35%
Answer : Percentage yield of ethanoic acid is 91% (to nearest whole number)
161
Chemistry Live! – Worked Solutions
W24.4
(i) X = Ethanol (C2 H5 OH)
Y = Propan-2-ol (C3 H5 OH)
(ii)
Relative molecular mass Na2 Cr2 O7 .2H2 O = 2(23) + 2(52) + 7(16) + 2(18) = 298
Relative molecular mass CH3 CHO = 12 + 3(1) + 12 + 1+ 16 = 44
Relative molecular mass CH3 COCH3 = 12 + 3(1) + 12 + 16 + 12 + 3(1) = 58
Mass
11.92
Number of moles of sodium dichromate = ––––––––––––––––– = ––––– = 0.04
Rel molecular mass
298
Sodium dichromate is the limiting reactant in both reactions, so we calculate the percentage
yields of ethanal and propanone on the amount of sodium dichromate present.
Group A Reaction
3C2 H5 OH + Cr2 O7 2- + 8H+ → 3CH3 CHO + 2Cr3+ + 7H2 O
1 mole → 3 moles
⇒ 0.04 mole → 0.04 × 3 mole
= 0.12 mole
= 0.12 × 44
= 5.28 g
i.e. the theoretical yield of ethanal is 5.28 g
Actual yield of product
Percentage yield of Ethanal = –––––––––––––––––––––––
Theoretical yield of product
2.75
= ––––– × 100
5.28
= 52% (to nearest whole number)
162
×
100
Chemistry Live! – Worked Solutions
Group B Reaction
3 C3 H5 OH + Cr2 O7 2- + 8H+ → 3CH3 COCH3 + 2Cr3+ + 7H2 O
1 mole →3 moles
⇒ 0.04moles → 0.04 × 3 moles
= 0.12 moles
= 0.12 × 58
= 6.96 g
i.e. the theoretical yield of propanone is 6.96 g
Actual yield of product
–––––––––––––––––––––––– × 100
Theoretical yield of product
Percentage yield of propanone =
5.15
=  × 100
6.96
= 74% (to nearest whole number)
Answer: (a) Percentage yield of ethanal is 52 %.
(b) Percentage yield of propanone is 74 %
W24.5
(i) 20% w/v solution of NaOH = 20 g of NaOH in 100 cm3 solution
(a) 100 cm3 = 20 g
⇒
30
30 cm3 = 20 × ----100
= 6 g of NaOH
(b) Relative molecular mass NaOH = 23 + 16 + 1 = 40
Mass in 30 cm3 solution
6
Number of moles of NaOH =  =  = 0.15
in 30 cm3 solution
Rel molecular mass
40
(ii) Rel molecular mass CH3 COOC2 H5 = 12 + 3(1) + 12 + 2(16) + 2(12) + 5(1) = 88
Mass of ethyl ethanoate = density × volume 0.9 × 6.6
5.94
mass
5.94
Number of moles of ethyl ethanoate =  =  = 0.0675
Rel molecular mass
88
Number of moles of NaOH = 0.15
163
Chemistry Live! – Worked Solutions
CH3 COOC2 H5
1 mole
⇒ 0.0675 mole
+
NaOH → CH3 COOH + C2 H5 OH
1 mole
0.0675 mole
0.0675 mole of sodium hydroxide would react with 0.0675 molesof ethyl ethanoate. However,
there is 0.15 mole of sodium hydroxide present, i.e. the sodium hydroxide is present in excess.
(iii) Relative molecular mass CH3 COOH = 12 + 3(1) +12 +2(16) +1 = 60
The ethyl ethanoate is the limiting reactant, so we calculate the percentage yield of ethanoic
acid on the amount of ethyl ethanoate present.
CH3 COOC2 H5 + NaOH → CH3 COOH + C2 H5 OH
1 mole → 1 mole
⇒ 0.0675 mole → 0.0675 mole
= 0.0675 × 60
= 4.05 g
i.e. the theoretical yield of ethanoic acid is 4.05 g
Percentage yield of ethanoic acid =
Actual yield of product

Theoretical yield of product
×
3.1
=  × 100
4.05
= 76.54%
Answer: (i) There are (a) 6 g and (b) 0.15 moles of NaOH in 30 cm3 solution
(ii) The sodium hydroxide is present in excess.
(iii) Percentage yield of ethanoic acid is 76.54%
164
100