(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${}^3$