Question

$60 mL$ of a mixture of $CO$ and $H _{2}$ yielded $20 mL$ of $CO _{2}$ after explosion with an excess of oxygen. What is the percentage of carbon monoxide in the mixture?

Answer 1

Vol. of mixture $\left( CO + H _{2}\right)=60 mL$

Vol. of $CO _{2}$ formed $=20 mL$

From reaction: $2 CO (g)+ O _{2}( g )=2 CO _{2}( g )$

2 vol. $\quad 1$ vol. 2 vol.

Vol. of $CO _{2}$ formed $=$ vol. of $CO$ reacted

$\therefore$ Vol. of $C O$ which reacted $=20 mL , \%$ of $C O$ in the mixture $=\frac{20}{60} \times 100=33.33 \%$.