(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;
(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;
(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1
(a) Balanced equation including N_2 from air
The balanced equation ignoring N_2 from air is
4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2
Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2
Including N_2 from air, the balanced equation is
4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2
(b) Balanced equation for 120 % stoichiometric combustion
Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2
Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2
Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2
The balanced equation is
4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2
(c) Minimum mass of air
Moles of O_2 required = 1700 kg C_11 H_7S
× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)
= 126.2 kmol O_2
Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2
Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)
× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2
Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air
(d) Air:fuel mass ratio for 100 % combustion
Air:fuel = 17 300 kg/1700 kg = 10.2 :1
(e) Air:fuel mass ratio for 120 % combustion
Mass of air = 17 300 kg × 1.20 = 20 760 kg air
Air:fuel = 20 760 kg/1700 kg = 12.2 :1
