A 20.0 cm3 mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature with excess of further contraction is found to be 13.0 cm3. A further contraction of 14.0 cm3 occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.

Sol. The reaction involved in the explosion process is : CO(g) + 1/2 O2 (g) → CO2 (g) x mL x/2 mL x mL CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l) y mL 2 y mL y mL The first step volume contraction can be calculated as : (x + x/2 + y + 2 y)-(x + y) = 13 ⇒ x + 4 y = 26 The second volume contraction is due to absorption of CO2. Hence, x + y = 14 …. (ii) Now, solving equations (i) and (ii), X = 10 mL, y = 4 mL and volume of He = 20 – 14 = 6 mL ⇒ Vol % of CO = 10/20 × 100 = 50% Vol of CH4¬ = 4/20 × 100 = 20% Vol % of He = 30%