SOLUTION : The volume of coal gas $=100 vol$.
$\begin{array}{ll}\text { and, therefore, } & \text { vol. of } H _{2}=45 vol . \\ & \text { vol. of } CH _{4}=30 vol . \\ \text { and } & \text { vol. of } CO =15 vol . \\ & \text { vol. of } C _{2} H _{2}=10 vol .\end{array}$
Reactions :
(i) $2 H _{2}( g )+ O _{2}( g )=2 H _{2} O ( l ) \quad \therefore \quad$ Vol. of $O _{2}$ reacted $=22 \cdot 5$ vol.
$\begin{array}{lll}2 \text { vol. } & 1 \text { vol. } & \text { zero vol. } \\ 45 \text { vol. } & \frac{45}{2} \text { vol. } & \text { zero vol. }\end{array}$
(ii) $CH _{4}( g )+2 O _{2}( g )= CO _{2}( g )+2 H _{2} O (l)$ Vol. of $O _{2}$ reacted $=60$ vol.
$\begin{array}{lllll}1 \text { vol. } & 2 \text { vol. } & 1 \text { vol. } & \text { nil } & \text { Vol. of } CO _{2} \text { formed }=30 vol . \\ 30 \text { vol. } & 60 \text { vol. } & 30 \text { vol. } & \text { ovol } & \end{array}$
(iii) $2 CO ( g )+ O _{2}( g )=2 CO _{2}( g ) \quad$ Vol. of $O _{2}$ reacted $=7 \cdot 5 vol$.
$\begin{array}{llll}2 \text { vol. } & 1 \text { vol. } & 2 \text { vol. } & \text { Vol. of } CO _{2} \text { formed }=15 vol . \\ 15 \text { vol. } & \frac{15}{2} \text { vol. } & 15 \text { vol. } & \end{array}$
(iv) $2 C _{2} H _{2}( g )+5 O _{2}( g )=4 CO _{2}( g )+2 H _{2} O ( l )$ Vol. of $O _{2}$ reacted $=25$ vol.
$\begin{array}{lllll}2 \text { vol. } & 5 \text { vol. } & 4 \text { vol. } & \text { ov. } \quad \text { Vol. of } CO _{2} \text { formed }=20 vol . \\ 10 \text { vol. } & 25 \text { vol. } & 20 vol . & \end{array}$
Now, total $O _{2}$ reacted $=(22 \cdot 5+60+7 \cdot 5+25)=115 \cdot 0$ vol.
Hence, vol. of unreacted $O_{2}=(200-115)$ vol. $=85$ vol.
And, vol. of $CO _{2}$ formed $=(30+15+20)$ vol. $=65$ vol.
Thus in the resulting mixture:
Total $CO _{2}$ present $= 6 5$ vol. Oxygen left unused $= 8 5$ vol.
Total volume $\quad=(65+85)$ vol. $= 1 5 0$ vol.
