Workbook – Worked Solutions Chapter 4 – The …chemistrylive2000.folensonline.ie/news/chemlive/news-38/Solutions... · Chemistry Live! – Worked Solutions 75 Workbook Chapter

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

75

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


76

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 = 80 g

1 mole of Br2 molecules = 80 × 2 = 160 g 0.125 moles of Br2 molecules = 160 × 0.125 = 20 g (c) 1 mole of H2O = (1 × 2) + 16 = 18 g 0.25 mole of H2O = 18 × 0.25 = 4.5 g (d) 1 mole of S atoms = 32 g 1 mole of S8 molecules = 32 × 8 = 256 g 0.5 moles of S8 molecules = 256 × 0.5 = 128 g (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 = 32 + (16 × 4) = 96 g 0.5 moles of SO4

2- ions = 96 × 0.5 = 48 g (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


77

W9.3 (a) Relative Molecular Mass NaOH = 23 + 16 + 1 = 40 i.e 1 mole of NaOH = 40 g

⇒ 4 moles of NaOH = 40 × 4 = 160 g

(b) Relative Molecular Mass CaCl2 = 40 + (2 × 35.5) = 40 + 71 = 111 i.e 1 mole of CaCl2 = 111g ⇒ 0.5 moles of CaCl2 = 111 × 0.5 = 55.5 g (c) Relative Molecular Mass Cl2 = 2 × 35.5 = 71 i.e. 1 mole of Cl2 = 71g ⇒ 2 moles of Cl2 = 71 × 2 = 142 g (d) Relative Molecular Mass MgO = 24 + 16 = 40 i.e. 1 mole of MgO = 40 g ⇒ 0.05 mole of MgO = 40 × 0.05 = 2 g (e) Relative Molecular Mass Fe2O3 = (2 × 56) + (3 × 16) = 112 + 48 = 160 i.e. 1 mole of Fe2O3 = 160 g ⇒ 3 moles of Fe2O3 = 160 × 3 = 480 g (f) Relative Molecular Mass C2H6 = (2 × 12) + (6 × 1) = 24 + 6 = 30 i.e. 1 mole of C2H6 = 30 g ⇒ 0.1 mole C2H6 = 30 × 0.1 = 3 g (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 H2O = (2 × 1) + 16 = 18 i.e. 1 mole of H2O = 18 g ⇒ 4 moles of H2O = 18 × 4 = 72 g


78

(j) Relative Molecular Mass Ca(NO3)2 = 40 + (14 × 2) + (16 × 6) = 40 + 28 + 96 = 164 i.e.1 mole of Ca(NO3)2 = 164 g ⇒ 0.075 moles of Ca (NO3)2 = 164 × 0.075 = 12.3 g (k) Relative Molecular Mass (NH4)2SO4 = (2 × 14) + (8 × 1) + 32 + (4 × 16) = 28 + 8 + 32 + 64 = 132 i.e. 1 mole of (NH4)2SO4 = 132 g ⇒ 0.02 moles of (NH4)2SO4 = 132 × 0.02 = 2.64 g (l) Relative Molecular Mass CuO = 63.5 + 16 = 79.5 i.e. 1 mole of CuO = 79.5 g ⇒ 0.25 moles of CuO = 79.5 × 0.25 = 19.875 g (m) Relative Molecular Mass SO3 = 32 + (3 × 16) = 32 + 48 = 80 i.e. 1 mole of SO3 = 80 g ⇒ 0.1 moles of SO3 = 0.1 × 80 = 8 g (n) Relative Molecular Mass Cu2O = (63.5 × 2) + 16 = 127 + 16 = 143 i.e.1 mole of Cu2O = 143 g ⇒ 0.25 moles of Cu2O = 143 × 0.25 = 35.75 g (o) Relative Molecular Mass (NH4)2CO3 = (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 = 2 × 80 = 160 i.e. 1 mole of Br2 = 160 g ⇒ 0.25 moles of Br2 = 160 × 0.25 = 40 g (q) Relative Molecular Mass H2SO4 = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 i.e. 1 mole of H2SO4 = 98 g ⇒ 0.8 moles of H2SO4 = 98 × 0.8 = 78.4 g


79

(r) Relative Molecular Mass FeCl2 = 56 + (2 × 35.5) = 56 + 71 = 127 i.e 1 mole of FeCl2 = 127 g ⇒ 4 moles of FeCl2 = 127 × 4 = 508 g (s) Relative Molecular Mass C12H26 = (12 × 12) + (26 × 1) = 144 + 26 = 170 i.e. 1 mole of C12H26 = 170 g ⇒ 1 × 10-4 moles of C12H26 = 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 H2O = (2 × 1) + 16 = 2 + 16 = 18 Mass 0.32 Number of moles of H2O = ––––––––––––––––– = ––––– = 0.0178 Rel molecular mass 18 (d) Relative Molecular Mass H2SO4 = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 Mass 441 Number of moles of H2SO4 = –––––––––––––––– = ––––– = 4.5 Rel molecular mass 98 (e) Relative Molecular Mass NH4NO3 = 14 + (4 × 1) + 14 + (3 × 16) = 14 + 4 + 14 + 48 = 80 Mass 10 Number of moles of NH4NO3 = –––––––––––––––– = ––– = 0.125 Rel molecular mass 80


80

(f) Relative Molecular Mass C6H12O6 = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 Mass 18 Number of moles of C6H12O6 = ––––––––––––––––– = –––– = 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 CuSO4.5H2O = 63.5 + 32 + (4 × 16) + (5 × 18) = 63.5 + 32 + 64 + 90 = 249.5 Mass 24.95 Number of moles of CuSO4.5H2O = ––––––––––––––––– = –––––– = 0.1 Rel molecular mass 249.5


81

W9.6 (a) 1 mole of H2O contains 6 × 1023 molecules 0.25 moles of H2O 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 H2SO4 contains 6 × 1023 molecules 0.001 mole of H2SO4 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


82

(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

27 × 2 × 1025 2 × 1025 atoms of Al = –––––––––––– g 6 × 1023

= 900 g


83

(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 H2O = (1 × 2) + 16 = 2 + 16 = 18

i.e. 1 mole of H2O = 18 g ⇒ 6 × 1023 molecules of H2O = 18 g

18 1 molecule of H2O = –––––––– g 6 × 1023

18 × 1.5 × 1025 1.5 × 1025 molecules of H2O = ––––––––––––– g 6 × 1023

= 450 g


84

(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

44 × 8 × 1021 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 Na2SO3 = (2 × 23) + 32 + (3 × 16) = 46 + 32 + 48 = 126 1 mole of Na2SO3 contains 6 × 1023 SO3

2- ions i.e. 126 g of Na2SO3 contain 6 × 1023 SO3

2- ions 6 × 1023 1 g of Na2SO3 contains –––––––– SO3

2- ions 126 2.1 × 6 × 1023 2.1 g of Na2SO3 contain –––––––––––– SO3

2- ions 126

= 1 × 1022 SO32- 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


85

(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.


86

(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 H2O.

1 mole of water = 18 g ⇒ 0.5 mole of water = 0.5 × 18 = 9 g


87

Workbook Chapter 10 - Properties of Gases

W10.1 Given s.t.p. V1 = 30 cm3 V2 = ? p1 = 975 kPa p2 =100 kPa T1 = 25 °C + 273 = 298 K T2 = 273 K p1V1 p2V2 ––––– = ––––– T1 T2

975 × 30 100 × V2 –––––––– = –––––––– 298 273 975 × 30 × 273 ⇒ V2 = –––––––––––––– 298 × 100 = 267.96 cm3 W10.2 Given s.t.p. V1 = 2.5 L V2 = ? p1 = 300 kPa p2 =100 kPa T1 = T2 p1V1 p2V2

––––– = ––––– T1 T2

300 × 2.5 = 100 × V2 300 × 2.5 ⇒ V2 = –––––––––– 100 = 7.5 L


88

W10.3 Old New V1 = 1.2 m3 V2 = ? p1 = 101 kPa p2 = 108 kPa T1 = 5 °C + 273 = 278 K T2 = 30 °C + 273 = 303 K . p1V1 p2V2 ––––– = ––––– T1 T2

101 × 1.2 108 × V2 –––––––––– = –––––––––– 278 303 101 × 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 = 7 L (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


89

(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


90

(f) Rel molecular mass H2S = (2 × 1) + 32 = 2 + 32 = 34 i.e. 1 mole of H2S = 34 g 34 g of H2S occupies 22.4 L at s.t.p 22.4 1 g of H2S occupies ––––– L 34 22.4 × 13 ⇒ 13 g of H2S 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


91

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)


92

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)


93

W10.9 Laboratory s.t.p. V1 = 328 cm3 V2 = ? p1 = 9.8 × 104 Pa p2 = 1 × 105 Pa

T1 = 100 °C + 273 = 373 K T2 = 273 K p 1V1 p2V2 ––––– = ––––– 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 1 cm3 of the vapour has a mass of ––––––– g 235.26 22,400 × 1.43 ⇒ 22,400 cm3 of the vapour have a mass of ––––––––––––––– 235.26 = 136.16 g i.e Relative molecular mass = 136 (to nearest whole number)


94

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)


95

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.


96

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 new V1 = 500 cm3 V2 = ? p1 = 1.01 × 105 Nm-2 p2 = 2.02 × 105 Nm-2 T1 = 273 K T2 = 819 K

p 1V1 p2V2

––––– = ––––– T1 T2

1.01 × 105 × 500 2.02 × 105 × V2 ––––––––––––––––– = ––––––––––––––– 273 819

1.01 × 105 × 500 × 819 ⇒ V2 = –––––––––––––––––––––––– 2.02 × 105 × 273 = 750 cm3


97

(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 Mass (e) Density = ––––––––– Volume Mass of one mole Mass of one mole 2.5 = –––––––––––––––––––– = ––––––––––––––––––– 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


98

Workbook Chapter 11 - Stoichiometry

W11.1 (a) Relative molecular mass Na2CO3.10H2O

= (23 × 2) + 12 + (16 × 3) + 10 [(1 × 2) + 16] = 46 + 12 + 48 + (10 × 18) = 46 +12 + 48 + 180 = 286

180 % H2O = ––––– × 100 = 62.94 %

286

(b) Relative molecular mass MgSO4.7H2O = 24 + 32 + (16 × 4) + 7 [(1 × 2) + 16] = 24 + 32 + 64 + (7 × 18) = 24 + 32 + 64 + 126 = 246 126

% H2O = ––––– × 100 = 51.22 % 246

W11.2 Relative molecular mass Fe2O3 = (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 Fe3O4 contains the higher % of Fe. W11.3

Element Percentage Percentage ––––––––––––––– Rel. At. Mass

Simplest ratio (divide by 1.35)

Carbon

64.9 64.9 –––––– = 5.41 12

5.41 ––––– = 4 1.35

Hydrogen 13.5 13.5 ––––– = 13.5 1

13.5 ––––– = 10 1.35

Oxygen 21.6 21.6 –––––– = 1.35 16

1.35 ––––– = 1 1.35

i.e. Empirical formula = C4H10O


99

Molecular Formula = empirical formula × n 74 = (C4H10O) × n 74 = [(12 × 4) + (1 × 10) + 16] n 74 = (48 + 10 + 16) n 74 = 74 n n = 1 ⇒ Molecular Formula = (C4H10O) × 1 = C4H10O i.e. Molecular Formula = empirical formula = C4H10O W11.4



Carbon

26.7 26.7 ––––– = 2.23 12

2.23 ––––– = 1 2.2

Hydrogen 2.2 2.2 –––– = 2.2 1

2.2 ––––– = 1 2.2

Oxygen 71.1 71.1 ––––– = 4.44 16

4.44 ––––– = 2 2.2

i.e. Empirical formula = CHO2 Molecular Formula = empirical formula × n 90 = (CHO2) × n 90 = [12 + 1 + (16 × 2)] n 90 = (12 + 1 + 32) n 90 = 45n n = 2 ⇒ Molecular Formula = (CHO2) × 2 = C2H2O4 i.e. Molecular Formula = C2H2O4


100

W11.5



Carbon

64.8 64.8 ––––– = 5.4 12

5.4 –––––– = 4 1.35

Hydrogen 13.6 13.6 ––––– = 13.6 1

13.6 ––––– = 10 1.35

Oxygen 21.6 21.6 –––––– = 1.35 16

1.35 ––––– = 1 1.35

i.e. Empirical formula = C4H10O

Molecular Formula = empirical formula × n 74 = (C4H10O) × n 74 = [(12 × 4) + (1 × 10) + 16] n 74 = (48 + 10 + 16) n 74 = 74n n = 1

⇒ Molecular Formula = (C4H10O) × 1 = C4H10O i.e. Molecular Formula = C4H10O


101

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


102

W11.7 CuSO4.xH2O → CuSO4 + x H2O 3.94 g 2.52 g mass of CuSO4 = 2.52 g mass of CuSO4.xH2O = 3.94 g mass of H2O in CuSO4.xH2O = 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 H2O = (1 × 2) + 16 = 18 mass number of moles of CuSO4 in CuSO4.xH2O = –––––––––––––– Rel. mol. mass 2.52 = –––––– = 0.0158 159.5 mass number of moles of water in CuSO4.xH2O = –––––––––––––––– Rel. mol. mass 1.42 = ––––– = 0.0789 18 Ratio of CuSO4 : H2O = 0.0158 : 0.0789 = 1 : 5

i.e. CuSO4.5H2O is the formula of the hydrated salt.


103

W11.8 M + oxygen → M2O 9.76 g 20.9 g mass of M = 9.76 g mass of M2O = 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 M2O 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 Mg3N2 = (24 × 3) + (14 × 2) = 72 + 28 = 100

3Mg + N2 → Mg3N2 3Mg → Mg3N2 (shortened version of equation) 3 moles → 1 mole 3 (24) g → 100 g

i.e. 72 g Mg → 100 g

100 1 g Mg → –––– g Mg3N2

72 33 × 100

33 g Mg → –––––––– g Mg3N2 72

= 45.83 g Mg3N2


104

W11.10 Relative molecular mass Fe2O3 = (56 × 2) + (16 × 3) = 112 + 48 = 160 2Al + Fe2O3 → Al2O3 + 2Fe Fe2O3 → 2Fe (shortened version of equation) 1 mole → 2 moles 160 g → 2 (56) g 160 g Fe2O3 → 112 g Fe 112 1 g Fe2O3 → –––– g Fe 160 30 × 112 30 g Fe2O3 → g Fe

160

= 21 g Fe W11.11 Relative mol. mass NH4NO3 = 14 + (1 × 4) + 14 + (16 ×3) = 14 + 4 +14 + 48 = 80 NH4NO3 → N2O + 2H2O NH4NO3 → N2O (shortened version of equation) 1 mole → 1 mole i.e. 80 g NH4NO3 → 22.4 L of N2O at s.t.p. 80 g NH4NO3 → 22,400 cm3 of N2O 80

–––––– g NH4NO3 → 1cm3 of N2O 22,400 550 × 80 g NH4NO3 → 550 cm3 of N2O 22,400 = 1.96 g NH4NO3


105

W11.12 Rel. molecular mass CaC2 = 40 + (12 × 2) = 40 + 24 = 64

CaC2 + 2H2O → C2H2 + Ca(OH)2 CaC2 → C2H2 (shortened version of equation) 1 mole → 1 mole

i.e. 64 g CaC2 → 22.4 L of C2H2 at s.t.p. 64 g CaC2 → 22,400 cm3 of C2H2 64

g CaC2 → 1cm3 of C2H2 22,400 350 × 64 g CaC2 → 350cm3 of C2H2 22,400 = 1 g CaC2

W11.13

Relative mol. mass NH4Cl = 14 + (1 × 4) + 35.5 = 14 + 4 + 35.5 = 53.5 2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O

2NH4Cl → 2NH3 (shortened version of equation) 2 moles → 2 moles

2 (53.5) g → 2 (22.4) L of NH3 at s.t.p. i.e. 107 g NH4Cl → 44.8 L of NH3

107 g NH4Cl → 44,800 cm3 of NH3

107 g NH4Cl →1cm3 of NH3 44,800 107 × 800 g NH4Cl → 800 cm3 of NH3

44,800

= 1.91 g NH4Cl


106

W11.14 Relative mol. mass KClO3 = 39 + 35.5 + (16 × 3) = 39 + 35.5 + 48 = 122.5 2KClO3 → 2KCl + 3O2

2KClO3 → 3O2 (shortened version of equation) 2 moles → 3 moles 2 (122.5) g → 3 (22.4) L of O2 at s.t.p i.e. 245 g KClO3 → 67.2 L of O2

245 g KClO3 → 67,200 cm3 of O2

245 g KClO3 → 1cm3 of O2 67,200

700 × 245 g KClO3 → 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) Cu2S copper (I) sulfate


107

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

Na2SO3 + 2HCl → 2NaCl + SO2 + H2O

(i) Relative mol. mass Na2SO3 = (23 × 2) + 32 + (16 × 3) = 46 + 32 + 48 = 126

mass number of moles of Na2SO3 =

Rel. mol. Mass

6.3 = = 0.05 126 (ii) Na2SO3 → SO2 (shortened version of equation)

1 mole → 1 mole 1 mole → 22.4 L 0.05 mole of Na2SO3 → (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


108

W11.18 4FeS2 + 11O2 → 2Fe2O3 + 8SO2

(i) Rel. mol. mass FeS2 = 56 + (32 × 2) = 56 + 64 = 120

Rel. mol. mass Fe2O3 = (56 × 2) + (16 × 3) = 112 + 48 = 160

4FeS2 → 2Fe2O3 (shortened version of equation) 4 moles → 2 moles 4(120) g → 2(160) g i.e. 480 g FeS2 → 320 g Fe2O3

320 1 g FeS2 → g Fe2O3

480

9.6 × 320 9.6 g FeS2 →

480

= 6.4 g Fe2O3 (ii) 4FeS2 → 8SO2 (shortened version of equation) 4 moles → 8 moles 4(120) g → 8 (22.4) L

480 g FeS2 → 179.2 L of SO2 at s.t.p.

179.2 1 g FeS2 → L of SO2

480

9.6 × 179.2 9.6 g FeS2 → of SO2

480 = 3.584 L of SO2


109

(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 3.584 × 6 × 1023 3.584 L of SO2 contains molecules 22.4

= 9.6 × 1022 molecules

W11.19 (a)

Element Percentage Percentage ––––––––––––– Rel. At. Mass


Carbon

38.7 38.7 ––––– = 3.23 12

3.23 –––– = 1 3.23

Hydrogen 9.68 9.68 ––––– = 9.68 1

9.68 –––– = 3 3.23

Oxygen 100 – (38.7 + 9.68) = 100 – 48.38 = 51.62

51.62 ––––– = 3.23 16

3.23 –––– = 1 3.23

i.e. Empirical formula = CH3O (b) Molecular Formula = empirical formula × n 62 = (CH3O) × n 62 = [12 + (1 × 3) + 16] n 62 = (12 + 3 + 16) n 62 = 31n n = 2 ⇒ Molecular Formula = (CH3O) × 2 = C2H6O2 i.e. Molecular Formula = C2H6O2


110

W11.20 MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O (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 MnO2 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 MnO2 → (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


111

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


112

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 cm3 of solution → M solution

56

= 1.79 M

(c) Rel. mol mass H2SO4 = 2(1) + 32 + 4(16)

= 2 + 32 + 64 = 98 98 g H2SO4 in 1L of solution → 1M solution

1 1g H2SO4 in 1L of solution → M solution 98 22 22 g H2SO4 in 1L of solution → M solution 98 22 × 10 22 g H2SO4 in 100 cm3 of solution → M solution

98

= 2.24 M


113

(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 Na2CO3 = 2(23) + 12 + 3(16)

= 46 + 12 + 48 = 106 106 g Na2CO3 in 1L of solution → 1M solution

1 1 g Na2CO3 in 1L of solution → M solution 106 12.5 12.5 g Na2CO3 in 1L of solution → M solution 106 12.5 × 5 12.5 g Na2CO3 in 200 cm3 of solution → M solution 106 = 0.59 M


114

W13.4 80 mg /100 cm3 = 0.08 g /100 cm3 Rel. mol mass C2H5OH = 2(12) + 5(1) + 16 +1

= 24 + 5 + 16 + 1 = 46 46 g C2H5OH in 1L of solution → 1M solution

1 1 g C2H5OH in 1L of solution → M solution 46 0.08 0.08 g C2H5OH in 1L of solution → M solution 46 0.08 × 10 0.08 g C2H5OH 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


115

(c) Volume × Molarity Number of moles of NH3 = 1000 750 × 0.12 = 1000 = 0.09 moles = 0.09 × 17 (Mr NH3 = 14 + 3(1) = 17) = 1.53 g (d) Volume × Molarity Number of moles of HCl = 1000 2000 × 2 = 1000 = 4 moles = 4 × 36.5 (Rel. mol mass HCl = 1 + 35.5 = 36.5) = 146 g (e) Volume × Molarity Number of moles of H2SO4 = 1000 20 × 0.1 = 1000 = 2 × 10-3 moles = 2 × 10-3 × 98 (Mr H2SO4 = 2(1) + 32 + 4(16) = 98) = 0.196


116

W13.6 Mg + 2HCl → MgCl2 + H2

Volume × Molarity Number of moles of HCl = 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 + H2O + 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


117

W13.8 Given: Vol dil = ? Vol dil × M dil Vol conc × M conc M dil = 0.01 M = Vol conc = 20 cm3 1000 1000 M conc = 1 M Vol dil × 0.01 = 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 Vol dil × M dil Vol conc × M conc Mdil = 2 M = Volconc = ? 1000 1000 Mconc = 16 M 2000 × 2 = Vol conc × 16

2,000 × 2 ⇒ Vol conc =

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 Vol dil × M dil Vol conc × M conc Mdil = 20 volume = Volconc = ? 1000 1000 Mconc = 100 volume 2,500 × 20 = Vol conc × 100

2,500 × 20 ⇒ Vol conc =

100

= 500 cm3

i.e. 500 cm3 of 100 volume H2O2 would have to be diluted to 2.5 L.


118

W13.12 Balanced equation: Ca(OH)2 + 2HCl → CaCl2 + 2H2O

Given: HCl Ca(OH)2 Va = 15.2 cm3 Ma = 0.05 M Va × Ma Vb × Mb na = 2 = Vb =25 cm3 na nb Mb = ? nb = 1 15.2 × 0.05 25 × Mb –––––––––– = –––––––

2 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: H2C2O4 + 2NaOH → Na2C2O4 + 2H2O Given: H2C2O4 NaOH Va = 18.7 cm3 Ma = ? Va × Ma Vb × Mb na = 2 –––––––– = –––––––– Vb =25 cm3 na nb Mb = 0.12 M nb = 1 18.7 × Ma 25 × 0.12 ––––––––– = –––––––––

1 2

25 × 0.12 ⇒ Ma = –––––––––– 2 ×18.7

= 0.08 moles / L = 0.08 × 90 ( Mr H2C2O4 = 90)

= 7.2 g / L

Answers: (a) 0.08 moles/L (b) 7.2 g/L


119

W13.14 Balanced equation: CH3COOH + NaOH → CH3COONa + H2O Given: CH3COOH NaOH Va = 20.25 cm3 Ma = ? Va × Ma Vb × Mb na = 1 –––––––– = –––––––– Vb =25 cm3 na nb Mb = 0.15 M nb = 1 20.25 × Ma 25 × 0.15 –––––––––– = –––––––––

1 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 CH3COOH = 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 %


120

W13.15 Balanced equation: CH3COOH + NaOH → CH3COONa + H2O Given: CH3COOH NaOH Va = 12.35 cm3 Ma = ? Va × Ma Vb × Mb na = 1 –––––––– = ––––––– Vb =25 cm3 na nb Mb = 0.12 M nb = 1 12.35 × Ma 25 × 0.12 ––––––––– = ––––––––

1 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 CH3COOH = 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 %


121

W13.16 (a) Balanced equation: H2C2O4 + 2NaOH → Na2C2O4 + 2H2O + CO2 Given: H2C2O4 NaOH Va = 20 cm3 Ma = ? Va × Ma Vb × Mb na = 1 ––––––– = ––––––– Vb =22.22 cm3 na nb Mb = 0.09 M nb = 2 20 × Ma 22.22 × 0.09 ––––––– = –––––––––––

1 2

22.22 × 0.09 ⇒ Mb = ––––––––––– 2 × 20

= 0.05 moles/L

1 mole H2C2O4 = 2(1) + 2(12) + 4(16) = 2 + 24 + 64 = 90 g

⇒ 0.05 moles H2C2O4 = 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 H2C2O4xH2O. Note from the above that by dissolving 6.3 g of the crystals of H2C2O4xH2O in 1 L of solution, we obtain a solution which contains 0.05 moles. Therefore, 0.05 moles H2C2O4.xH2O = 6.3 g

6.3 1 mole H2C2O4.xH2O = –––– = 126 0.05

i.e. Rel. mol mass of H2C2O4.xH2O = 126 ⇒ 2(1) + 2(12) + 4(16) + x(2 +16) = 126 90 + 18x = 126 18x = 36 x = 2 i.e. formula is H2C2O4.2H2O Answer: (a) % Water of crystallisation = 28.57 % (b) x = 2


122

W13.17 Balanced equation: H2C2O4 + 2NaOH → Na2C2O4 + 2H2O + CO2 Given: H2C2O4 NaOH Va = 25 cm3 Ma = ? Va × Ma Vb × Mb na = 1 ––––––– = ––––––– Vb =22.75 cm3 na nb Mb = 0.35 M nb = 2 25 × Ma 22.75 × 0.35 –––––––– = –––––––––––

1 2 22.75 × 0.35

⇒ Mb = ––––––––––– 2 × 25 = 0.1593 moles/L

0.1593 moles H2C2O4.xH2O per litre = (0.1593 / 4) moles H2C2O4.xH2O / 250cm3

= 0.0398 moles H2C2O4.xH2O/250 cm3 Note from the above that by dissolving 5 g of the crystals of H2C2O4xH2O in 250 cm3 of solution, we obtain a solution which contains 0.0398 moles. Therefore, 0.0398 moles H2C2O4.xH2O = 5 g

5 1 mole H2C2O4.xH2O = –––––– = 125.63 0.0398

i.e. Rel. mol mass of H2C2O4.xH2O = 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 H2C2O4.2H2O Answer: x = 2 Yes – he results obtained by both students are consistent.


123

Workbook Chapter 15 – Volumetric Analysis: Oxidation W15.1 Balanced equation: MnO4

- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O Given:- MnO4

- Fe2+ Vo =28.5 cm3 Mo = ? Vo × Mo Vr × Mred no = 1 ––––––– = –––––––– Vr =25 cm3 no nr Mred = 0.11 M nr = 5 28.5 × Mo 25 × 0.11 ––––––––– = ––––––––

1 5

25 × 0.11 ⇒ Mo = ––––––––– 5 × 28.5

= 0.0193 moles/L = 0.0193 × 158 (Mr KMnO4 = 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+ + 4H2O Given: MnO4

- Fe2+ Vo =22.5 cm3 Mo = 0.02 M Vo × Mo Vr × Mred no = 1 ––––––– = ––––––––– Vred =25 cm3 no nr Mr = ? nr = 5 22.5 × 0.02 25 × Mred –––––––––– = –––––––––

1 5

22.5 × 0.02 × 5 ⇒ Mred = ––––––––––––– 25

= 0.09 moles/L = 0.09 × 392 (Mr (NH4)2 (SO4)FeSO4.6H2O

= 392) = 35.28 g / L

Answer: (a) 0.09 moles/L (b) 35.28 g/L


124

W15.3 (a) Balanced equation: MnO4


- Fe2+ Vo =23.5 cm3 Mo = 0.025 M Vo × Mo Vr × Mre no = 1 –––––––– = ––––––– Vr =25 cm3 no nr Mred = ? nr = 5 23.5 × 0.025 25 × Mred ––––––––––– = –––––––––

1 5 23.5 × 0.025 × 5

⇒ Mred = –––––––––––––– 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%


125

W15.4 (b) Balanced equation: MnO4


- Fe2+ Vo = 19.65 cm3 Mo = 0.024 M Vo × Mo Vr × Mred no = 1 –––––––– = ––––––––– Vred = 25 cm3 no nr Mr = ? nr = 5 19.65 × 0.024 25 × Mred –––––––––––– = ––––––––

1 5 19.65 × 0.024 × 5

⇒ Mred = ––––––––––––––– 25 = 0.0943 moles/L

0.0943 moles FeSO4.xH2O / L = (0.0943 / 4) moles FeSO4.xH2O / 250 cm3 = 0.0236 moles FeSO4.xH2O / 250 cm3 i.e. by dissolving 5.71 g of the crystals of FeSO4.xH2O in 250 cm3 of solution, we obtain a solution which contains 0.0236 moles. Therefore, 0.0236 moles FeSO4.xH2O = 5.71 g

5.71 1 mole FeSO4.xH2O = –––––– = 241.95 0.0236 ⇒ 56 + 32 + 4(16) + x(2 + 16) = 241.95 152 + 18x = 241.95 18x = 89.95 x = 5 i.e. formula is FeSO4.5H2O Answer: (b) x = 5


126

W15.5 (ii) 392 g of (NH4)2(SO4)FeSO4.6H2O in 1 L of solution → 1 M solution

1 1 g of (NH4)2(SO4)FeSO4.6H2O in 1 L of solution → ––– M solution 392 1 × 11.76 11.76 g of (NH4)2(SO4)FeSO4.6H2O in 1 L of solution → –––––––––– M solution 392 4 × 11.76 10.52 g of (NH4)2(SO4)FeSO4.6H2O in 250 cm3 of solution → –––––––––– M solution 392 = 0.12 M (v) Balanced equation: MnO4


- Fe2+ Vo =33.3 cm3 Mo = ? Vo × Mo Vr × Mred no = 1 –––––––– = –––––––– Vred =25 cm3 no nr Mr = 0.12 M nr = 5 33.3 × Mo 25 × 0.12 –––––––– = –––––––––

1 5 25 × 0.12

⇒ Mo = ––––––––– 5 × 33.3 = 0.018 moles/L

Answer: (ii) 0.12 M, (v) 0.018 M


127

W15.6 (vi) (a) Balanced equation: MnO4


- Fe2+ Vo = 22.5 cm3 Mo = 0.02 M Vo × Mo Vr × Mred no = 1 ––––––– = ––––––––– Vr =25 cm3 no nr Mred = ? nr = 5 22.5 × 0.02 25 × Mred ––––––––––– = ––––––––

1 5 22.5 × 0.02 × 5

⇒ Mr = ––––––––––––– 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. xH2O / 250 cm3

⇒ mass of H2O / 250 cm3 = 8.82 – 6.39 = 2.43 g mass of water % water of crystallisation = ––––––––––––––––––––––––––––– × 100 mass of (NH4)2 (SO4)FeSO4. xH2O 2.43 = ––––– × 100 = 27.55 %

8.82 0.09 moles (NH4)2 (SO4)FeSO4.xH2O/L = (0.09/4) moles (NH4)2 (SO4)FeSO4.xH2O / 250 cm3 = 0.0225 moles (NH4)2 (SO4)FeSO4.xH2O/250 cm3 i.e. by dissolving 8.82 g of the crystals of (NH4)2 SO4FeSO4.xH2O in 250 cm3 of solution, we obtain a solution which contains 0.0225 moles. Therefore, 0.0225 moles (NH4)2 SO4FeSO4.xH2O = 8.82 g


128

8.82

1 mole (NH4)2 SO4FeSO4.xH2O = –––––– = 392 0.0225 ⇒ 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 SO4FeSO4.6H2O Answer: (vi) (a) 0.09 M (b) 25.56 g/L, 27.55%, x = 6. W15.7 (iii) (a) Balanced equation: MnO4


- Fe2+ Vo = 5.47 cm3 Mo = 0.015 M Vo × Mo Vr × Mred no = 1 ––––––– = –––––––– Vr = 20 cm3 no nr Mred = ? nr = 5 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


129

(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


- Fe2+ Vo = 5.75 cm3 Mo = 0.015 M Vo × Mo Vr × Mred no = 1 ––––––––– = ––––––––– Vr =25 cm3 no nr Mred = ? nr = 5 5.75 × 0.015 25 × Mr ––––––––––– = ––––––––

1 5 5.75 × 0.015 × 5

⇒ Mr = –––––––––––––– 25


130

= 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 (b) 0.048 g (c) 53.75%


131

W15.9 (a) Balanced equation:

I2 + 2S2O32- → S4O6

2- + 2I- Given:- I2

S2O32-

Vo = 25 cm3 Mo = ? Vo × Mo Vr × Mred no = 1 –––––––– = –––––––– Vr =22.15 cm3 no nr Mred = 0.12 M nr = 2 25 × Mo 22.15 × 0.12 –––––––– = ––––––––––––

1 2 22.15 × 0.12

⇒ Mo = –––––––––– 2 × 25

= 0.0532 moles/L = 0.0532 × 254 = 13.51 g/L

Answer: (a) 0.0532 moles/L, (b) 13.51 g/L


132

W15.10 (a) 2MnO4

- + 10I- + 16H+ → 2Mn2+ + 5I2 + 8H2O I2 + 2S2O3

2- → S4O62- + 2I-

From the balanced equations, we see that 2 moles of MnO4- produce five moles of I2 and these 5

moles of I2 would then react with 10 moles of S2O32-.

i.e. 2MnO4- = 5I2 = 10S2O3

2- Given: MnO4

- S2O32-

Vo = 25 cm3 Mo = 0.022 M Vo × Mo Vr × Mr no = 2 ––––––– = ––––––– Vr = 22.45 cm3 no nr Mr = ? nr = 10 25 × 0.022 22.45 × Mr –––––––––– = ––––––––––

2 10 25 × 0.022 × 10

⇒ Mr = –––––––––––––– 2 × 22.45 = 0.1225 moles/L

= 0.1225 × 248 = 30.38 g/L

(Rel. molecular mass Na2S2O3.5H2O = 248) Answer: (a) 0.1225 moles/L, (b) 30.38 g/L


133

W15.11 (vi) (a) Balanced equation:

2S2O32- + I2 → S4O6

2- + 2I- Given: I2

S2O32-

Vo = 25 cm3 Mo = 0.05 M Vo × Mo V × Mred no = 1 ––––––– = –––––––– Vr = 31.2 + 31.3 = 31.25 cm3 no nr 2 Mred = ? 25 × 0.05 31.25 × Mred nr = 2 –––––––––– = ––––––––––

1 2 25 × 0.05 × 2

⇒ Mred = –––––––––––– 31.25

= 0.08 moles/L = 0.08 × 248 = 19.84 g / L

(Rel. mol mass Na2S2O3.5H2O = 248) Concentration of Na2S2O3.5H2O in g / L = number of moles /L × rel. mol mass (vii) 19.84 g Na2S2O3.5H2O / L = (19.84 / 2) g Na2S2O3.5H2O / 500 cm3 = 9.92 g Na2S2O3.5H2O / 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%


134

W15.12 (a) ClO - + 2I- + 2H+ → Cl- + I2 + H2O 2S2O3

2- + I2 → S4O62- + 2I-

From 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 S2O3

2-. i.e. 1ClO- = 1I2 = 2S2O3

2- Given:- ClO - S2O3

2- Vo = 25 cm3 Mo = ? Vo × Mo Vr × Mred no = 1 ––––––– = –––––––– Vr =19.56 cm3 no nr Mred = 0.25 M nr = 2 25 × Mo 19.56 × 0.25 –––––––– = –––––––––––

1 2 19.56 × 0.25

⇒ Mo = ––––––––––– 2 × 25 = 0.0978 moles / L

⇒ Concentration of original solution = 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%


135

Workbook Chapter 16 – Rates of Reaction W16.3 (ii) graph (iv) From graph: 1 24 cm3 of 0.2 M Na2S2O3 → –– = 0.01 s-1

t ⇒ t = 100 seconds W16.4 (iii) graph

(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

mass number of moles of O2 = –––––––––––– 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


136

(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

2H2O2 → 2H2O + O2

2H2O2 → O2 (shortened version of equation) 2 moles → 1 mole ⇒ 2 × 0.015 → 0.015 moles = 0.03 moles H2O2

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

From graph: 44.8 cm3 are liberated at a time of 1.2 minutes

(vi) (a) 50 – 6.5 Instantaneous rate of reaction at 0.5 minutes = ––––––– = 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


137

Workbook Chapter 17 – Chemical Equilibrium W17.7 CH3COOH + C2H5OH º CH3COOC2H5 + H2O Initially: 1 4 0 0 At Equil. : 1 – x 4 – x x x Conc. at equil. [1 – x / V] [4 – x / V] [x / V] [x / V] V is the volume of the container which we are not given in the question. [CH3COOC2H5] [H2O] (x / V) (x / V) Kc = –––––––––––––––––––– = –––––––––––––––––– = 4 [CH3COOH] [C2H5OH] (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 = 0.93 6 6


138

Chemically, 5.74 does not make sense since we started with only 1 mole of CH3COOH and 4 moles of C2H5OH. Therefore, we take the value of x = 0.93.

Therefore equilibrium concentrations are :-

CH3COOH = 1 – x = 1 – 0.93 = 0.07 mol/L C2H5OH = 4 – x = 4 - 0.93 = 3.07 mol/L CH3COOC2H5 = x = 0.93 mol/L H2O = x = 0.93 mol/L Answer: Equilibrium concentration of CH3COOC2H5 = 0.93 mol/L Equilibrium concentration of H2O = 0.93 mol/L W17.8 2HI ↔ H2 + I2 Initially: 0.1 0 0 At Equil. : 1 – 0.02 0.01 0.01 = 0.08 Conc. at equil.: [0.08] [0.01] [0.01]

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


139

W17.9 Rel molecular mass CH3COOH = 12 + 3(1) + 12 + 2(16) + 1 = 12 + 3 + 12 + 32 + 1 = 60 Mass 36 Initial number of moles of CH3COOH = ––––––––––––––––– = ––– = 0.6 Rel molecular mass 60 Mass 12 At equil. number of moles of CH3COOH = –––––––––––––––– = ––– = 0.2 Rel molecular mass 60 Rel molecular mass C2H5OH = 2(12) + 5(1) + 16 + 1 = 24 + 5 + 16 +1 = 46 Mass 27.6 Initial number of moles of C2H5OH = ––––––––––––––––– = ––––– = 0.6 Rel molecular mass 46 CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O Initially: 0.6 0.6 0 0 At Equil. : 0.6 – 0.4 0.6 – 0.4 0.4 0.4 = 0.2 = 0.2 Conc. at equil. [0.2 / V] [0.2 / V] [0.4 / V] [0.4 / V] V is the volume of the container which we are not given in the question. [CH3COOC2H5 ] [H2O] (0.4 / V) (0.4 / V) Kc = ––––––––––––––––––– = ––––––––––––––– [CH3COOH] [C2H5OH] (0.2 / V) (0.2 / V) The V terms cancel out in the Kc expression (0.4) (0.4) ⇒ Kc = –––––––– = 4 (0.2) (0.2) Mass 60 Initial number of moles of CH3COOH = –––––––––––––––– = ––– = 1.0 Rel molecular mass 60 Rel molecular mass CH3COOC2H5 = 12 + 3(1) + 12 + 2(16) + 2(12) + 5(1)


140

= 12 + 3 + 12 + 32 + 24 + 5 = 88 Mass 70.4 At equil. number of moles of CH3COOC2H5 = ––––––––––––––––– = –––– = 0.8 Rel molecular mass 88 CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O Initially: 1 C 0 0 At Equil. : 1 – 0.8 C – 0.8 0.8 0.8 = 0.2 Conc. at equil. [0.2 / V] [C – 0.8 / V] [0.8 / V] [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. [CH3COOC2H5] [H2O] (0.8 / V) (0.8 / V) Kc = ––––––––––––––––––– = ––––––––––––––––– [CH3COOH] [C2H5OH] (0.2 / V) (C – 0.8 / V) The V terms cancel out in the Kc expression (0.8) (0.8) ⇒ Kc = ––––––––––– = 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


141

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) + NO2 (g) ↔ SO3(g) + NO(g) Initially: 0.12 0.1 0 0 At Equil. : 0.12 – 0.06 0.1 – 0.06 0.06 0.06 = 0.06 = 0.04

Conc. at equil. [0.06 / V] [0.04 / V] [0.06 / V] [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) ⇒ Kc = ––––––––––– = 1.5 (0.06) (0.04)

Answer Kc = 1.5


142

W17.11 2HI ↔ H2 + I2

Initially: 0 2 1

At equil.: 2x 2 - x 1 - x

Conc. at equil. : [2x] [2 – x] [1 –x] (V =1 L)

[H2][ I2] (2 – x) (1 – x) Kc = –––––––– = ––––––––––– = 0.02 [HI] 2 (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 or x = 1.72 = 0.93 1.84 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


143

W17.12 N2O4(g) ↔ 2NO2(g) Initially: 0.1 0 At Equil. : 0.1 - x 2x Conc. at equil. : [0.1 - x] [2x] (V = 1 L) [NO2] 2 (2x) 2 Kc = ––––––– = –––––––– = 0.36 [N2O4] (0.1 - x) 4x2 ⇒ –––––– = 0.36 0.1 - x

cross multiply ⇒ 4x2 = 0.36 (0.1 - x) ⇒ 4x2 = 0.036 – 0.36x ⇒ 4x2 + 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 or x = -1.2 = -0.15 8 8

Chemically, -0.15 does not make sense, therefore, we take the value of x = 0.06.

Therefore equilibrium concentrations are:- N2O4 = 0.1 – x = 0.1 – 0.06 = 0.04 mol/L NO2 = 2x = 2 (0.06) = 0.12 mol/L

Answer Equilibrium concentration of N2O4 = 0.04 mol/L Equilibrium concentration of NO2 = 0.12 mol/L


144

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


145

W18.4 (a) H2SO4 → 2H+ + SO4

2- 1 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+ + Cl- 1 mole → 1 mole ⇒ 0.2 mole → 0.2 mole pH = - log 10 [H+] = -log 10 (0.2) = 0.70 (c) HCl → H+ + Cl- 1 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+ + OH- 1 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 H2SO4 / 200 cm3 = (0.049 × 5) g H2SO4 / L = 0.245 g H2SO4 / L

Rel molecular mass H2SO4 = 2(1) + 32 + 4(16) = 98 Mass in 1 L 0.245 Number of moles of H2SO4 / L = –––––––––––––––– = ––––– = 0.0025 Rel molecular mass 98 H2SO4 → 2H+ + SO4

2- 1 mole → 2 mole ⇒ 0.0025 mole → 0.0025 × 2 mole

= 0.005 mole pH = - log 10 [H+] = -log 10 (0.005) = 2.30


146

W18.5 (a) NaOH → Na+ + OH- 1 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+ + OH- 1 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+ + OH- 1 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+ + OH- 1 mole → 1 mole ⇒ 5 ×10-5 mole → 5 ×10-5 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+ + OH- 1 mole → 1 mole ⇒ 7.14 ×10-4 mole → 7.14 ×10-4 mole

pOH = - log 10 [OH-] = -log 10 (7.14 ×10-4) = 3.15 pH = 14 – pOH =14 –3.15 = 10.85


147

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-] ⇒ Kw = [H+]2

At 25 oC 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 oC 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 oC = 7 pH at 57 oC = 6.52


148

W18.9 8 g CH3COOH / 250 cm3 = (8 × 4) g CH3COOH / L = 32 g CH3COOH / L Rel molecular mass CH3COOH = 12 + 3(1) + 12 + 2(16) + 1 = 60 Mass in 1 L 32 Number of moles of CH3COOH / L = ––––––––––––––––– = ––– = 0.533

Rel molecular mass 60

CH3COOH → CH3COO- + 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 CH3COOH / 250 cm3 = (3 × 4) g CH3COOH / L = 12 g CH3COOH / L Rel molecular mass CH3COOH = 12 + 3(1) + 12 + 2(16) + 1 = 60 Mass in 1 L 12 Number of moles of CH3COOH / 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


149

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 ⇒ Macid = –––––––––– = ––––––––– = 1.25 ×10-3 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 H3BO3 / L

Rel molecular mass H3BO3 = 3(1) + 11 + 3(16) = 62

Mass in 1L 5.5 Number of moles of HNO3 / L = –––––––––––––––– = –––– = 0.089 Rel molecular mass 62

H3BO3 → H+ + H2BO3-

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 ⇒ Ka = 1.225 × 10-4

Answer: Ka = 1.225 × 10-4


150

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+ + A- 1 mole → 1 mole ⇒ 2 × 10-3 mole → 2 × 10-3 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 ⇒ Ka = 8 × 10-6 Answer: [HA] = 2 × 10-3 mol/L Ka = 8 × 10-6


151

Workbook Chapter 19 – Environmental Chemistry – Water

W19.2 (e) Given:- Ca2+ edta VCa = 50 cm3 MCa = ? VCa × MCa Ved × Med nCa = 1 –––––––––– = ––––––––– Ved = 18 cm3 nCa ned Med = 0.01 M ned = 1 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:- Ca2+ edta VCa = 50 cm3 MCa = ? VCa × MCa Ved × Med nCa = 1 ––––––––– = –––––––––– Ved = 11.5 cm3 nCa ned Med = 0.01 M ned = 1 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


152

(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.


153

W19.4 (iv) Given: Ca2+ edta VCa = 50 cm3 MCa = ? VCa × MCa Ved × Med nCa = 1 ––––––––– = ––––––––– Ved = 15 cm3 nCa ned Med = 0.01 M ned = 1 50 × MCa 15 × 0.01 ––––––––– = –––––––––

1 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:- Ca2+ edta VCa = 50 cm3 MCa = ? VCa × MCa Ved × Med nCa = 1 ––––––––– = ––––––––– Ved = 9.5 cm3 nCa ned Med = 0.01 M ned = 1 50 × MCa 9.5 × 0.01 –––––––– = ––––––––––

1 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


154

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:- O2 S2O3

2- Vo = 200 cm3 Mo = ? Vo × Mo Vr × Mr no = 1 –––––––– = –––––––– Vr = 22.7 cm3 no nr Mr = 0.01 M nr = 4 200 × Mo 22.7 × 0.01 –––––––– = ––––––––––

1 4 22.7 × 0.01

⇒ Mo = –––––––––––– 4 × 200 = 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.


155

After 5 days Given: O2 S2O3

2- Vo = 200 cm3 Mo = ? Vo × Mo Vr × Mr no = 1 ––––––––– = –––––––– Vr = 4.8 cm3 no nr Mr = 0.01 M nr = 4 200 × Mo 4.8 × 0.01 –––––––– = ––––––––––

1 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.


156

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 + H2O ∆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 + H2O Answer: Heat of neutralisation = - 54.6 k J mol-1


157

W21.14 Required Equation:- CH4 (g) + 4Cl2 (g) → CCl4 (l) + 4HCl (g) ∆H =?

(a) reversed CH4 (g) → C(s) + 2H2 (g) ∆H = 74.9 kJ mol-1

(b) C(s) + 2Cl2 (g) → CCl4 (l) ∆H = -139 kJ mol-1

(c) × 4 2H2 (g) + 2Cl2 (g) → 4HCl (g) ∆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: CH3OH (l) + 1½O 2 (g) → CO2 (g) + 2H2O(l) ∆H =?

(c) reversed CH3OH (l) → C(s) + 2H2 (g) + ½O 2 (g) ∆H = 250 kJ mol-1

(b) × 2 2H2 (g) + O2 (g) → 2H2O(l) ∆H = -572 kJ mol-1

(a) C(s) + O2 (g) → CO2 (g) ∆H = -394 kJ mol-1

CH3OH (l) + 2H2 (g) + O2 (g) + C(s) + O2 (g) → C(s) + 2H2 (g) + ½O 2 (g) + 2H2O(l) + CO2 (g)

i.e. CH3OH (l) + 1½O 2 (g) → CO2 (g) + 2H2O(l) ∆H = -716 kJ mol-1

Answer: ∆H = -716 kJ mol-1

W21.16 Required Equation:- 2CH4 (g) + O2 (g) →2CH3OH(l) ∆H =?

(a) × 2 2CH4 (g) + 2H2O (g) → 2CO (g) + 6H2 (g) ∆H = 412 kJ mol-1

(c) 2H2 (g) + O2 (g) → 2H2O (l) ∆H = -434 kJ mol-1

(b) × 2 4H2 (g) + 2CO (g) → 2CH3OH (g) ∆H = -256 kJ mol-1

2CH4 (g) + 2H2O (g) + 2H2 (g) + O2 (g) + 4H2 (g) + 2CO (g) → 2CO (g) + 6H2 (g)+2H2O (l) + 2CH3OH (g) i.e 2CH4 (g) + O2 (g) → 2CH3OH(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 2H2 (g) + O2 (g) → 2H2O (l) ∆H = -571.6 kJ mol-1

(c) C (s) + O2 (g) → CO2 (g) ∆H = -393.5 kJ mol-1

(a) reversed CO2 (g) + 2H2O (l) → CH4 (g) + 2O2 (g) ∆H = 890.4 kJ mol-1

2H2 (g) + O2 (g) + C (s) + O2 (g) + CO2 (g) + 2H2O (l) → 2H2O (l) + CO2 (g) + CH4 (g) + 2O2 (g)

i.e. C (s) + 2H2 (g) → CH4 (g) ∆H = -74.7 kJ mol-1



158

W21.18 Required Equation:- 3C (s) + 4H2 (g) → C3H8 (g) ∆H =?

(b) × 3 3C (s) + 3O2 (g) → 3CO2 (g) ∆H = -1,182 kJ mol-1

(c) × 4 4H2 (g) + 2O2 (g) → 4H2O (l) ∆H = -1,144 kJ mol-1

(a) reversed 3CO2 (g) + 4H2O (l) → C3H8 (g) + 5O2 (g) ∆H = 2,220 kJ mol-1

3C (s) + 3O2 (g) + 4H2 (g) + 2O2 (g) +3CO2 (g) + 4H2O (l) → 3CO2 (g) + 4H2O (l) + C3H8 (g) + 5O2 (g)

i.e. 3C (s) + 4H2 (g) → C3H8 (g) ∆H = -106 kJ mol-1 Answer: ∆H = -106 kJ mol-1 W21.19 Required Equation:- 2C (s) + H2 (g) → C2H2 (g) ∆H =?

(b) × 2 2C (s) + 2O2 (g) → 2CO2 (g) ∆H = -788 kJ mol-1

(c) H2 (g) + ½O2 (g) → H2O (l) ∆H = -286 kJ mol-1

(a) reversed 2CO2 (g) + H2O (l) → C2H2 (g) + 2½O 2 (g) ∆H = 1,299 kJ mol-1

2C (s) + 2O2 (g) + H2 (g) + ½O 2 (g) + 2CO2 (g) + H2O (l) → 2CO2 (g) + H2O (l) + C2H2 (g) + 2½O 2 (g

i.e. 2C (s) + H2 (g) → C2H2 (g) ∆H = 225 kJ mol-1

Answer: ∆H = 225 kJ mol-1

W21.20 Required Equation:- ½N 2 (g) + 1½H 2 (g) → NH3 (g) ∆H =?

(a) reversed × ¼ ½N 2 (g) + 1½H 2O (l) → NH3 (g) + ¾O 2 (g) ∆H = 317 kJ mol-1

(b) × 1½ 1½ H 2 (g) + ¾O 2 (g) → 1½H 2O (l) ∆H = -428.7 kJ mol-1

½N 2 (g) + 1½H 2O (l) +1½ H 2 (g) + ¾O 2 (g) → NH3 (g) + ¾O 2 (g) +1½H 2O (l)

i.e. ½N 2 (g) + 1½H 2 (g) → NH3 (g) ∆H = -111.7 kJ mol-1


W21.21 Required Equation:- Ca (s) +C (s) + 1½O 2 (g) → CaCO3 (s) ∆H =?

(a) × ½ Ca (s) + ½O 2 (g) + CO2 → CaCO3 (s) ∆H = -320 kJ mol-1

(b) C (s) + O2 (g) → CO2 (g) ∆H = -393 kJ mol-1

_________________

Ca (s) + ½O 2 (g) + CO2 + C (s) + O2 (g) → CaCO3 (s)+ CO2 (g)

i.e. Ca (s) +C (s) + 1½O 2 (g) → CaCO3 (s) ∆H = -713 kJ mol-1

Answer: ∆H = -713 kJ mol-1


159

W21.22 Required Equation:- C2H2 (g) + HCN (g) → C3H3N (g) ∆H =? Given (a) 2C (s) + H2 (g) → C2H2 (g) ∆H = 233 kJ mol-1

(b) ½H 2 (g) + C (s) + ½N 2 (g) → HCN (g) ∆H = 114.5 kJ mol-1

(c) 3C (s) + 1½H 2 (g) + ½N 2 (g) → C3H3N (g) ∆H = 194.5 kJ mol-1

(a) reversed C2H2 (g) → 2C (s) + H2 (g) ∆H = -233 kJ mol-

(b) reversed HCN (g→ ½H 2 (g) + C (s) + ½N 2 (g) ∆H = -114.5 kJ mol-1

(c) 3C (s) + 1½H 2 (g) + ½N 2 (g) → C3H3N (g) ∆H = 194.5 kJ mol-1

C2H2 (g) +HCN (g) + 3C (s) + 1½H 2 (g) + ½N 2 (g) → 2C (s) + H2 (g) + ½H 2 (g) + C (s) + ½N 2 (g) +C3H3N (g) i.e C2H2 (g) + HCN (g) → C3H3N (g) ∆H = -153 kJ mol-1 Answer: ∆H = -153 kJ mol-1 W21.23 Required Equation:- H2 (g) + S (s) + 2O2 (g) → H2SO4 (l) ∆H =? Given (a) H2 (g) + ½O 2 (g) → H2O (l) ∆H = -286 kJ mol-1

(b) S (s) + O2 (g) → SO2 (g) ∆H = -297 kJ mol-1

(c) 2SO2 (g) + O2 (g) → 2SO3 (g) ∆H = -196 kJ mol-1

(d) SO3 (g) + H2O (l) → H2SO4 (l) ∆H = -133 kJ mol-1

(a) H2 (g) + ½O 2 (g) → H2O (l) ∆H = -286 kJ mol-1

(b) S (s) + O2 (g) → SO2 (g) ∆H = -297 kJ mol-1

(c) × ½ SO2 (g) + ½O 2 (g) → SO3 (g) ∆H = -98 kJ mol-1

(d) SO3 (g) + H2O (l) → H2SO4 (l) ∆H = -133 kJ mol-1

H2 (g) + ½O 2 (g) + S (s) + O2 (g) + SO2 (g) + ½O 2 (g) + SO3 (g) + H2O (l) → H2O (l) + SO2 (g) + SO3 (g) +H2SO4 (l)

i.e. H2 (g) + S (s) + 2O2 (g) → H2SO4 (l) ∆H = -814 kJ mol-1 Answer: ∆H = -814 kJ mol-1


160

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 → 2Fe2O3 + 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 = –––––––––––––––––––––––– × 100 Theoretical yield of product

2725 = ––––––––– × 100 53,333.76

= 5.11%

Answer : Percentage yield of sulphur dioxide is 5.11%. W24.2 Relative molecular mass NH4Cl = 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

2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O 2 moles 1 mole ⇒ 0.37 moles 0.37 moles 2 = 0.185 moles


161

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. 2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O 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 C2H5OH = 2(12) + 5(1) + 16 + 1 = 46 Relative molecular mass CH3COOH = 12 + 3(1) +12 +2(16) +1 = 60 Mass 24 Number of moles of ethanol = –––––––––––––––– = –––– = 0.52 Rel molecular mass 46 3C2H5OH + 2Cr2O7

2- + 16H+ → 4Cr3+ + 3CH3COOH + 11H2O 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 Actual mass yield of product Percentage yield = ––––––––––––––––––––––––– × 100 Theoretical yield of product 28.5 = ––––– × 100 31.2 = 91.35% Answer : Percentage yield of ethanoic acid is 91% (to nearest whole number)


162

W24.4 (i) X = Ethanol (C2H5OH)

Y = Propan-2-ol (C3H5OH) (ii) Relative molecular mass Na2Cr2O7.2H2O = 2(23) + 2(52) + 7(16) + 2(18) = 298 Relative molecular mass CH3CHO = 12 + 3(1) + 12 + 1+ 16 = 44 Relative molecular mass CH3COCH3 = 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 3C2H5OH + Cr2O7

2- + 8H+ → 3CH3CHO + 2Cr3+ + 7H2O 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 = ––––––––––––––––––––––– × 100 Theoretical yield of product 2.75 = ––––– × 100 5.28 = 52% (to nearest whole number)


163

Group B Reaction 3 C3H5OH + Cr2O7

2- + 8H+ → 3CH3COCH3 + 2Cr3+ + 7H2O 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 Percentage yield of propanone = –––––––––––––––––––––––– × 100 Theoretical yield of product 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 CH3COOC2H5 = 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


164

CH3COOC2H5 + NaOH → CH3COOH + C2H5OH 1 mole 1 mole ⇒ 0.0675 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 CH3COOH = 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.

CH3COOC2H5 + NaOH → CH3COOH + C2H5OH 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 Actual yield of product Percentage yield of ethanoic acid = × 100 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%

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