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  3. Chemistry
  4. At 300 K and 1 atm , 15 mL of a gaseous hydrocarbon requires 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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Q. At $300 \,K$ and $1 \,atm , 15\, mL$ of a gaseous hydrocarbon requires $375\, mL$ 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 :

JEE MainJEE Main 2016Some Basic Concepts of Chemistry

Solution:

$\underset{15\, {\text{mL}}}{C_X H_Y} + \underset{15(x + \frac{y}{4}){\text{mL}}}{(x+\frac{Y}{4})O_2}\rightarrow \underset{15 \,{\text{ x mL}}}{xCO_2}+\frac{Y}{2}H_2O$

$V_{O_2} = \frac{20}{100} \times 375 = 75 \,mL = 15 (x + \frac{y}{4})$

$\Rightarrow \, x + \frac{y}{4} = 5 $

$\Rightarrow \, C_3H_8$