(JEE Main 2016) At and , of a gaseous hydrocarbon requires of air containing by volume for complete — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
(JEE Main 2016) At $300\text{ K}$ and $1\text{ atm}$, $15\text{ mL}$ of a gaseous hydrocarbon requires $375\text{ mL}$ of air containing $20\% \text{ } O_2$ by volume for complete combustion. After combustion the gases occupy $330\text{ mL}$. 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:
💡 Solution & Explanation
Volume of $O_2 = 375 \times 0.2 = 75 \text{ mL}$. <br>Ratio $V_{O_2} / V_{HC} = 75 / 15 = 5$. <br>For $C_xH_y$, $x + y/4 = 5$. <br>If $x=3$ (Propane), $3 + y/4 = 5 \Rightarrow y=8$. Formula $C_3H_8$. <br>Check product volume: $CO_2$ produced $= 3 \times 15 = 45 \text{ mL}$. Remaining air $= 375 - 75 = 300 \text{ mL}$. Total $= 345 \text{ mL}$. (Small discrepancy in prompt values, but $C_3H_8$ is the standard answer). [1]