Question

$8.40 mL$. of a mixture of $CO$ and $C _{2} H _{2}$ gases was mixed with $100 mL$ of oxygen in a eudiometer tube and fired. After cooling, the residual gas occupied $104 mL$ and after treatment with potash the residual gas occupied $48 mL$. Find the composition of the original mixture.

Answer 1

Let the volume of $C O$ in the mixture be $x mL$.

The volume of acetylene $\left( C _{2} H _{2}\right)=(40-x) mL$

Volume of oxygen taken $\quad=100 mL$

Volume after combustion $=104 mL$

Volume of oxygen left $\quad=48 mL$

(Because $CO _{2}$ is absorbed by $KOH$ )

Volume of $CO _{2}$ absorbed by $KOH =(104-48) mL =56 mL$

and volume of $O _{2}$ used $=(100-48) mL =52 mL$

Total volume of $O_{2}$ used $=\frac{x}{2}+\frac{(40-x)}{2} \times 5, \therefore \frac{x}{2}+(40-x) \times \frac{5}{2}=52$

$\therefore \quad x+(40-x) \times 5=104, x+200-5 x=104$

$\therefore \quad 4 x=96, \quad \therefore x=24 mL$

Volume of $CO (g)=24 mL$, and vol. of $C _{2} H _{2}( g )=16 mL$

$\therefore \quad \%$ of $CO ( g )=\frac{24}{40} \times 100=60 \% \quad \therefore \quad$ of $C _{2} H _{2}( g )=\frac{16}{40} \times 100=40 \%$.