Question

A $20.0cc$ mixture of $CO,C{H}_4$ and $He$ gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be $13.0cc$ A further contraction of $14.0cc$ occurs when the residual gas is treated with $KOH$ solution. The composition of the gaseous mixture in terms of volume percentage will be:

• A. $CO:40\%,C{H}_4:20\%,He:40\%$
• B. $CO:40\%,C{H}_4:30\%,He:20\%$
• C. $CO:50\%,C{H}_4:20\%,He:30\%$
• D. $CO:50\%,C{H}_4:10\%,He:40\%$

Answer 1

The correct option is C $CO:50\%,C{H}_4:20\%,He:30\%$

$\therefore x+y+7=20$ -----(1)

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

$x$ $\frac{x}{2}$ $0$

$0$ - $x$ $\Delta =\frac{x}{2}$

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

$xy$ $2y$ $0$ $0$

$0$ - $y$ $2y\downarrow$ $\Delta =2y$

No one with

Now,

$KOH$ reacts with $C{O}_2$

$\therefore x+y=14$ -------(2)

$\therefore z=6$

$\therefore z=\frac{6}{20}\times 100=30\%$

Contractor of $13$ earner

$\therefore \frac{x}{2}+2y=13$

$\Rightarrow x+4y=26$ -------(3)

$3y=12$ (from (2)&(3))

$\Rightarrow y=4$

$x=10$

$\%x=50\%,\%y=20\%$