Question

40 ml of mixture of $C_2{H}_2$ and CO is mixed with 100 ml of $O_2$ gas and the mixture is exploded. The residual gases occupied 104 ml and when these are passed through KOH solution, the volume becomes 48ml. All the volume are at same the temperature and pressure. Determine the composition of the original mixture.

• A. $C_2{H}_2=16ml,CO=24ml$
• B. $C_2{H}_2=23ml,CO=14ml$
• C. $C_2{H}_2=30ml,CO=26ml$
• D. $C_2{H}_2=33ml,CO=28ml$

Answer 1

The correct option is A $C_2{H}_2=16ml,CO=24ml$

Let the original mixture contains x ml of CO and (40-x) ml of $C_2{H}_2$.
Th balanced chemical equations are as shown below.
$\underset{40-x}{C_2{H}_2}+\underset{2.5(40-x)}{\frac{5}{2}{O}_2}\to \underset{2(40-x)}{2C{O}_2}+H_{2}O\left(l\right)$
$\underset{x}{C}O+\underset{0.5x}{\frac{1}{2}{O}_2}\to \underset{x}{C{O}_2}$
Total volume is 104 ml.
Hence, $2(40-x)+x+(100-2.5(40-x)-0.5x=104$.
On solving this, we get $x=24ml$.
Hence, the volumes of CO and $C_2{H}_2$ are 24 ml and $40-24=16$ ml respectively.