A mixture of $C H_{4}$ and $C_{2} H_{4}$ was completely burnt in excess of oxygen, yielding equal volumes of $C O_{2}$ and steam. Calculate the percentages of the compounds in the original mixture:

25%

C H_{4}

and 75%

C_{2} H_{4}

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30%

C H_{4}

and 70%

C_{2} H_{4}

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75%

C H_{4}

and 25%

C_{2} H_{4}

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53.33%

C H_{4}

and 46.67%

C_{2} H_{4}

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Solution

The correct option is D 53.33% $C H_{4}$ and 46.67% $C_{2} H_{4}$
we have,
$C H_{4} + O_{2} \to C O_{2} + 2 H_{2} O$
and, $C_{2} H_{4} + O_{2} \to 2 C O_{2} + 2 H_{2} O$
if they are producing equal volume of $C O_{2}$ gas, that means moles of $C H_{4}$ is double the moles of $C_{2} H_{4}$

let, moles of $C H_{4}$ = 1, and mass becomes = 16 g
and moles of $C_{2} H_{4}$ = 0.5 and mass becomes = 14 g
so, percentage composition of $C_{2} H_{4}$ = $\frac{14}{30} * 100$ = 46.6 %
percentage composition of $C H_{4}$ = $\frac{16}{30} * 100$ = 53.33%

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