(a) (i) X - oxygen
Y - sulphur

(ii) I. - S(s) + O\(_{2(g)}\) \(\to\) SO\(_{2(g)}\) (1)
II. - 2SO\(_{2(g)}\) + O\(_{2(g)}\) \(\to\) 2SO\(_{3(g)}\) (2)
III. - SO\(_{3(g)}\) + H\(_2\)SO\(_{4(aq)}\) \(\to\) H\(_2\)S\(_2\)O\(_7\)\(_{(l)}\) (1)
IV. - H\(_2\)S\(_2\)O\(_7\)\(_{(l)}\) + H\(_2\)O\(_{(l)}\) 2H\(_2\)SO\(_{4(aq)}\) (2)

(iii) Vanadium (V) oxide (1)

(iv) It is because the reaction is exothermic
(v) - In bleaching wood pulp, wool silk, hay
- as germicide and fumigant/ disinfectant
- as refrigerant / refrigerating gas/coolant
- as preservative for grains and fruits
- manufacture of H\(_2\)SO\(_4\)
- as a reducing agent in metallurgy
- for dechlorination


(b) From the equation,
2 mol of H\(_2\) reacts with 1 mol of O\(_2\)
hence 40 cm\(^3\) of H\(_2\) would react with 20 cm\(^3\) of O\(_2\)
Volume of unused oxygen = (30 – 20) cm\(^3\)
= 10 cm\(^3\)

(c) (i) CaCO\(_3\) \(\to\) CaO + CO\(_2\)

(ii) CaCO\(_3\) = 40 + 12 + (16 x 3) = 100 g
CaO = 40 + 16 = 56 g
100 g of CaCO\(_3\) \(\to\) 56 g CaO
1.0 g will give \(\frac{1}{100}\) x 56 = 0.56 g

(iii) 1 mole of a gas at s.t.p. 22.4 dm\(^{-3}\)
1 mole of CaCO\(_3\) produces 1 mole of CO\(_2\)
100 g of CaCO\(_3\) produces = 22.4 dm\(_3\) of CO\(_2\)
1.0 g of CaCO\(_3\) will produce = \(\frac{1}{100}\) x 22.4
= 0.224 dm\(^3\) of CO\(_2\)
(iv) Volume at 15\(^o\)C and 760 mm Hg
Temp = 273 + 15\(^o\)C = 288
\(\frac{V_1}{T_2} = \frac{V_2}{T_2}\)

Volume V\(_2\) = 0.224 x \(\frac{288}{273}\) = 0. 236 dm\(^3\)