Section Six : Important Industrial Products

Suggested Answer

1. (a) (i) Chlorine gas. 1

(ii) Turn colourless potassium bromide solution brown. 1,1

(b) OCl^- + Cl^- + 2H⁺ ¾® Cl₂ + H₂O 1

(d) (i) Anode: 2Cl^- ¾® Cl₂ + 2e^- Cathode: 2H⁺ + 2e^- ¾® H₂ 2

(ii) From the above equation: the mole ratio of H₂ and Cl₂ = 1 : 1 1

Mole ratio of gas is equal to volume ratio, vol. of Cl₂ = 100cm³ 1

(iii) Sodium hydroxide solution. 1

(e) (i) Turn colourless potassium bromide solution brown. 1

Cl₂ + 2Br^- ¾® 2Cl^- + Br₂ 1

(ii) Red colour fades 1

ClO^- + dye ¾® Cl^- + (dye+O) 1

2. (a) ClO^- decomposes under heat / sunlight. 1

2ClO^- ¾® 2Cl^- + O₂1

(b) Keep away from acid. 1

Bleach reacts with acid and forms toxic chlorine gas. 1

ClO^- + Cl^- + 2H⁺ ¾® Cl₂ + H₂O

3. (a) no. of mole of S = 8000/32 = 250 1

1 mole S can produce 1 mole sulphuric acid.

no. of mole of H₂SO₄ = 250

mass of sulphuric acid produce = 250 x 98 = 24500g = 24.5kg 1

(b) Stage II is a reversible reaction. 1

The actual amount of SO₃ formed is less than the theoretical value. 1

(ii) Platinum / vanadium(V) oxide. 1

(iii) Sulphur dioxide and air should be purified to remove impurities. 1

(d) If sulphur trioxide dissolves in water directly, large amount of heat produced which will cause a mist of concentrated acid droplets. 1,1

(e) B, D 1

(f) (C₆H₁₀H₅)_n ¾® 6nC + 5nH₂O 1

Carbon is formed as concentrated sulphuric acid is a dehydrating agent.

(g) Turn orange acidified potassium dichromate solution green. 1

3SO₂ + Cr₂O₇^2- + 2H⁺ ¾® 3SO₄^2- + 2Cr³⁺ + H₂O 1

[Accept any other possible oxidizing agent. E.g. Br₂, Fe³⁺ / MnO₄^-/H⁺ ……]