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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.

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.

Grade:11

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
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%

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