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Stoichiometric Calculations

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Question

A mixture of $20$ mL $C O, C H_{4}$ and $N_{2}$ was burnt in excess of $O_{2}$ , resulting in reduction of $13$ mL of volume. The residual gas was then treated with $K O H$ solution to show a contraction of $14$ mL in volume. The amount of volume (in mL) of $N_{2}$ in the mixture is: (assume $H_{2} O$ is in the liquid state and all measurements are made at constant P and T)

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Solution

Let the mixture contain $a, b$ and $c$ mL of $C O, C H_{4}$ and $N_{2}$ in it. Therefore, $a + b + c = 20 . . . (i)$
The reactions during combustion of mixture are as follows:
$C O + \frac{1}{2} O_{2} \to C O_{2}$
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O (l)$
$N_{2} + O_{2} \to N o r e a c t i o n$
Thus, reduction in volume is $2 b$ due to 2nd reaction and $\frac{a}{2}$ due to 1st reaction.
Total $C O_{2}$ formed $= a + b =$ volume absorbed by KOH
$∴ a + b = 14 . . . (i i)$
and, $(a / 2) + 2 b$ $=$ $13 . . . (i i i)$
By equations $(i), (i i) a n d (i i i)$ , $a = 10 m L, b = 4 m L, c = 6 m L$

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