Question

A $20.0 \,cm ^{3}$ mixture of $CO , CH _{4}$ and $He$ gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be $13.0 \,cm ^{3} . A$ further contraction of $14.0 \,cm ^{3}$ occurs when the residual gas is treated with $KOH$ solution. Find out the composition of the gaseous mixture in terms of volume percentage.

Answer 1

The reaction involved in the explosion process is
$\underset{x \,mL}{CO(g)} + \underset{\frac{x}{2}mL}{\frac{1}{2}O_2(g)} \rightarrow \underset{x\,mL}{CO_2(g)}$
$\underset{y\,mL}{CH_4(g) } + \underset{2y\,mL}{2O_2(g)} \rightarrow \underset{y\,mL}{CO_2(g)} + 2H_2O(l)$
The first step volume contraction can be calculated as :
$\Rightarrow \left(x+\frac{x}{2}+y+2 y\right)-(x+y) =13 $
$x+4 y =26 ....$ (i)
The second volume contraction is due to absorption of $CO _{2}$.
Hence, $x+y=14 .....$ (ii)
Now, solving equations (i) and (ii), $x=10 \,mL , y=4 \,mL$ and volume of $He =20-14=6 \,mL$
$\Rightarrow Vol \% $ of $ CO =\frac{10}{20} \times 100=50 \% $
$Vol \% $ of $ CH _{4}=\frac{4}{20} \times 100=20 \% $
$Vol \% $ of $ He =30 \%$