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Percentage Composition

Fifty millili...

Question

Fifty millilitre of a mixture of $C O$ and $C H_{4}$ , was exploded with $85 m l$ of $O_{2}$ . The volume of $C O_{2}$ , produced was $50 m l$ , Calculate the percentage composition of the gaseous mixture.

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Solution

Let the $v o l u m e o f C O = x m l$
$V o l u m e o f C H_{4}, = (50 - x) m l$
$C O x m l + \frac{1}{2} O_{2} \frac{x}{2} m l \to C O_{2} x m l . . . (i)$
$C H_{4} (50 - x) + 2 O_{2} 2 (50 - x) \to C O_{2} (50 - x) m l + 2 H_{2} O . . . (i i)$
$T o t a l v o l u m e o f O_{2} = \frac{x}{2} + 2 (50 - x) = 85$
$∴ x = 10 m l$
$V o l u m e o f C O = 10 m l$
$V o l u m e o f C H_{4} = (50 — 10) = 40 m l$
$P e r c e n t a g e o f C O = \frac{10 \times 100}{50} = 20 %$
$P e r c e n t a g e o f C H_{4} = (100 — 20) = 80 %$

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