of a gaseous hydrocarbon undergoes explosion with an excess of oxygen. Upon cooling to room temperat — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
$v\text{ mL}$ of a gaseous hydrocarbon undergoes explosion with an excess of oxygen. Upon cooling to room temperature, a volume contraction of $2.5v\text{ mL}$ is observed. When the residual gases are treated with concentrated aqueous $KOH$, a further volume contraction of $2v\text{ mL}$ occurs. Deduce the formula of the hydrocarbon.
💡 Solution & Explanation
Step 1: Combustion equation: $C_xH_y(g) + (x+y/4)O_2(g) \rightarrow xCO_2(g) + (y/2)H_2O(l)$. Step 2: Second contraction is due to $CO_2$ absorption by $KOH$. Volume of $CO_2 = v \cdot x = 2v \implies x = 2$. Step 3: First contraction = (Vol of reactants) - (Vol of gaseous products). At room temperature, $H_2O$ is liquid (negligible volume). Contraction = $[v + v(x + y/4)] - v \cdot x = v(1 + x + y/4 - x) = v(1 + y/4)$. Step 4: Equate to given contraction: $v(1 + y/4) = 2.5v \implies 1 + y/4 = 2.5 \implies y/4 = 1.5 \implies y = 6$. The hydrocarbon is $C_2H_6$.