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Measurement of Volume

The

Question

The % by volume of $C_{4} H_{10}$ in a gaseous mixture of $C_{4} H_{10}, C H_{4}$ and $C O$ is $40$ . When $200 m l$ of the mixture is burnt in excess of $O_{2}$ . Find volume (in ml) of $C O_{2}$ produced.

220

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340

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440

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560

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Solution

The correct option is D $440$
$%$ by volume of $C_{4} H_{10}$ in a gaseous mixture $= 40 %$

Volume of $C_{4} H_{10}$ in the mixture $= 40 %$ of $200 m l = 80 m l$
Combustion reaction of $C_{4} H_{10}$
$C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} + 5 H_{2} O$

Volume of $C O_{2}$ produced by combustion of $C_{4} H_{10} = 4 \times 80 = 320 m l$

Combustion reaction of $C H_{4}$
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

Combustion reaction of $C O$
$2 C O + O_{2} \to 2 C O_{2}$

Volume of $C H_{4} + C O$ in mixture = 200- 40=120 ml

Equal volume of $C O_{2}$ is produced in each $C H_{4}$ and $C O$ combustion.
Volume of $C O_{2}$ produced by the Combustion reaction of $C O$ and $C H_{4}$ = 120 ml

Total volume of $C O_{2} = 320 + 120 = 440 m l$

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