At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requries 375 mL air containing 20% $O_{2}$ by volume for complete combustion. After combustion the gases occupy 330 mL. 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:

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Solution

The correct option is A

$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$

So, volume of $O_{2}$ required for the combustion of 1 mL hydrocarbon = 5 mL.

So, volume of $O_{2}$ required for the combustion of 15 mL of hydrocarbon = 75 mL (i.e., 20% of 375 mL air)

NOTE : But for this, the total volume of gases after combustion should be 345 mL, rather than 330 mL.

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