At $300 K$ and $1 a t m$ , $15 m L$ of a gaseous hydrocarbon requires $375 m L$ air containing $20 %$ $O_{2}$ by volume for complete combustion. After combustion the gases occupy $330 m L$ . Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:

C_{3} H_{8}

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C_{4} H_{8}

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C_{4} H_{10}

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C_{3} H_{6}

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Solution

The correct option is A $C_{3} H_{8}$
375 ml of 20% oxygen corresponds to $375 \times \frac{20}{100} = 75$ mL.

The ratio of the volume of oxygen used for combustion to volume of hydrocabon $= \frac{75}{15} = \frac{5}{1}$ .

This ratio can be obtained with propane.

$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$
Water is in the liquid form.

15 ml of propane on combustion will give $3 \times 15 = 45$ mL of carbon dioxide.

$375 - 75 = 300$ ml of air (except oxygen) will remain.

Total volume will be $300 + 45 = 345$ .

Some carbon dioxide will dissolve in liquid water. Hence, the total volume will be less than 345 mL. This explains a volume of 330 mL.

Hence, the correct option is $A$

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