One litre mixture of $C O$ and $C O_{2}$ is passed through red-hot charcoal in a tube. The new volume becomes 1.4 L. Fine out the composition of the mixture by volume.

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Solution

The equation of the reaction is:
$C O_{2} + C \to 2 C O$
According to the equation, when 1 litre of $C O_{2}$ reacts with 1 mol of coke, two litres of CO is formed. Our mixture of gases have volume 1.4 L. Thus it is composed of CO and non-reacted $C O_{2}$ .
If x litres of $C O_{2}$ is reacted, 2x litres of CO is formed. Total volume of mixture is the sum of $C O_{2}$ and CO volumes:
V(mixture) = V(new volume $C O_{2}$ )+V(CO) = 1.4 L
V(new volume $C O_{2}$ ) = V(initial volume $C O_{2}$ ) –V(reacted $C O_{2}$ ) = (1-x)L
V(CO) = 2x
So, we have equation:
1.4 =1-x+2x
The solution is x = 0.4 L
Thus, the mixture consists of 1-0.4 = 0.6 L of $C O_{2}$ and 2 $\times$ 0.4 = 0.8 L of CO(iii)
The composition of product is 0.6 ltr of $C O_{2}$ and 0.8 ltr of CO.

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