One litre of a mixture of CO and `CO_2` is passed through red-hot charcoal. The volume now becomes 1.6 litre. Find the composition of the mixture by volume.

Explanation

When the gaseous mixture is passed over red hot charcoal, carbon dioxide is converted to carbon monoxide according to the equation
`underset("1 vol.")(CO_(2))+underset(("Charcoal"))(C)rarrunderset("2 vol.")(2CO)`
Thus, the increase in volume is because of conversion of `CO_(2)` gas to CO gas,
Let the volume of `CO_(2)` in the mixture = xL
`:.` volume of CO in the mixture` = (1-x)L`
Now, 1 L of `CO_(2)` form CO = 2L
`:. x L` of `CO_(2)` form CO = 2x L
The total volume of gaseous mixture after reaction `=2x +1 -x=(1+x)L`
But the volume actually recorded `= 1.6 L :. 1 +x =1.6 or x = 0.6 L`
`:.` Volume of `CO_(2)` in the mixture = 0.6 L
Volume of CO in the mixture `= (1- 0.60)=0.4 L`.