Question

A $20.0$ ${cm}^3$ mixture of $CO$, $C{H}_4$ and $He$ gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume of contraction is found to be $13$ ${cm}^3$. A further contraction of $14$ ${cm}^3$ occurs when the residual gas is treated with $KOH$ solution. The composition of $CO$ in the mixture in terms of volume percentage is:

Answer 1

Let volume of $CO,C{H}_4$ and $He$ be $a,b$ and $c$ respectively.

Given, $\therefore a+b+c=20...\left(i\right)$

The reactions are:

$CO+\frac{1}{2}{O}_2\to C{O}_2$

$C{H}_4+2O_2\to C{O}_2+2H_{2}O$

$\therefore$ Volume of $C{O}_2$ formed $(a+b)$ $=$ $14...\left(ii\right)$

By Eqs. (i) and (ii) , $c=6$

Also, an initial volume of the mixture -final volume after sparking $=13$

$a+b+c+V_{O_2}-[a+b+c+V_{O_2}]=13$

$\therefore V_{O_2}$ used $=13$

or, $\frac{a}{2}+2b=13$

or, $a+4b=26...\left(iii\right)$

$\therefore a=10,b=4$

$\therefore$ % of $CO$ $=\frac{10}{20}\times 100=50$%

% of $C{H}_4=\frac{4}{20}\times 100=20$%

% of $He$ =$\frac{6}{20}\times 100=30$%