Question

10ml of mixture of CO,CH_{​​​​​4} and N_{​​​​​​2} exploded with excess of oxygen gave a contraction of 6.5ml.There was a further contraction of 7 ml when the residual gas was treated with KOH.What is the amount of CH_{​​​​​4} in the original mixture?

Answer 1

Let the volumes of CO, CH4 and N2 in the mixture be x ml, y ml and (10-x-y) ml respectively. We ignore the volume of water formed in the combustion as it is a liquid and its volume is small. Nitrogen combusts only at very very high temperatures. So it remains unreactive.

2 CO + O2 ==> 2 CO2 CH4 + 2 O2 ==> CO2 + 2 H2O
x ml x/2 ml x ml y ml 2 y ml y ml (liquid)

Volume of oxygen used: 2y+x/2 ml
Total volume of all gases before combustion = 10 + 2 y + x/2 ml

After combustion the total volume is: 10 ml
as CO2 : x + y N2: 10 - x- y

Reduction in volume: 10 + 2y + x/2 - 10 = 2 y + x/2 = 6.5 ml
=> 4 y + x = 13 ml --- (1)

When the mixture of CO2 + N2 passes over KOH, all of CO2 is absorbed. Only N2 remains. Then reduction of 7 ml corresponds to that of CO2.
=> x + y = 7 ml --- (2)

Solving the two equations, x = 5 ml and y = 2 ml

Volumes of CO = 5 ml, CH4 = 2 ml , N2 = 3 ml.

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