Question

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 :

• A. $C_3H_6$
• B. $C_3H_8$
• C. $C_4H_8$
• D. $C_4H_{10}$

Answer 1

Answer: (B)

$C_3H_8$

Answer 2

$\underset{15\, {\text{mL}}}{C_X H_Y} + \underset{15(x + \frac{y}{4}){\text{mL}}}{(x+\frac{Y}{4})O_2}\rightarrow \underset{15 \,{\text{ x mL}}}{xCO_2}+\frac{Y}{2}H_2O$
$V_{O_2} = \frac{20}{100} \times 375 = 75 \,mL = 15 (x + \frac{y}{4})$
$\Rightarrow \, x + \frac{y}{4} = 5$
$\Rightarrow \, C_3H_8$