SOLUTION : Vol. of $\left( CO + CH _{4}+ C _{2} H _{6}\right)=10 mL$
Vol. of $O_{2}$ added $\quad=40 mL$
Vol. of $CO _{2}$ formed $=12 mL$
Vol. of $O _{2}$ unreacted $=23 mL$
$\therefore$ Initial volume $=(10+40) mL \quad=50 mL$
Final volume ( after reaction with $\left.O_{2}\right)=(23+12) mL =35 mL$
$\therefore$ Contraction in volume $\quad=(50-35) mL =15 mL$
and vol. of $O _{2}$ reacted $=(40-23) mL =17 mL$
The mixture contains $CO , CH _{4}$ and $C _{2} H _{6}$ and all three react with oxygen. Therefore, the values of three unknowns $x, y$ and $z$ are to be worked out, when
$x=\text { vol. of } CO , y=\text { vol. of } CH _{4} \text { and } z=\text { vol. of } C_{2} H _{6} \text {. }$
In order to find out these values at least three equations are to be framed and we have data, viz, vol. of mixture, vol. of $O _{2}$ reacted, vol. of $CO _{2}$ formed and contraction in volume. We proceed in the following way:
Reactions :
$z mL \quad \frac{7 z}{2} mL \quad 2 z mL$ oml Contraction in vol. $=\frac{5 z}{2} mL$
Now: $\quad x+y+z=10$
Total $CO _{2}: x+y+2 z=12$
Total $O_{2}$ reacted : $\frac{x}{2}+2 y+\frac{7 z}{2}=17 \quad$ or, $x+4 y+7 z=34$
Total contraction : $\frac{x}{2}+2 y+\frac{5 z}{2}=15 \quad$ or, $x+4 y+5 z=30$
For arriving at the values of $x, y$ and $z$, we use only (I), (II) and (III) and then confirm that these data correspond to the equation (IV).
Form (I) and (II), subtracting them,
$z=2 mL \text { (vol. of } C _{2} H _{6} \text { ) }$
(This value of $z$ is also obtained from equations III and IV.)
Subtracting (II) from (III), $3 y+5 z=22$
Putting the value of $z, \quad 3 y+10=22,3 y=12, y=4 mL$ (vol. of $CH _{4}$ )
From (I), $\quad x+4+2=10$
$\therefore \quad x=4 mL ($ vol. of $C O)$
$\therefore$ Vol. of $CO =4 mL$, Vol. of $CH _{4}=4 mL$, Vol. of $C _{2} H _{6}=2 mL$.
