Question

$10 mL$ of a mixture of $CO _{2} CH _{4}$ and nitrogen exploded with excess of oxygen gave a contraction of $6.5 mL$, there was a further contraction of $7 mL$ when the residual gas was treated with caustic potash. What was the composition of the original mixture?

Answer 1

Given that

$\begin{array}{ll}\text { Vol. of }\left( CO + CH _{4}+ N _{2}\right) & =10 mL \\ \text { Contraction in vol. after reaction } & =6.5 mL \\ \text { Contraction in vol. due to absorption by } KOH & =7 mL \end{array}$

$\therefore$ Vol. of $CO _{2}$ formed $=7 mL$

We suppose that vol. of  $CO$ in the mixture $=x mL$

and vol. of  $CH _{4}  =y mL$

Note : (We do not consider the unknown term of $N_{2}$ as it does not take part in such reactions). For solution, only two equations are required and they can easily be deduced by. knowing the contraction in vol. and vol. of $CO _{2}$ formed by the use of unknown $x$ and $y$.)

Reactions :

Again,

Total contraction : $\frac{x}{2}+2 y=6 \cdot 5 \ldots$ (I) Total $CO _{2}: x+y=7 \cdot 0$

Solving (I) and (II), $\quad x=5 mL ($ vol. of $C O) ; \quad y=2 mL$ (vol. of $CH _{4}$ )

and vol. of $N_{2}=10-(5+2)=3 mL$ (vol. of $\left.N_{2}\right)$.