$5 ext{ mL}$ of a gas containing only carbon and hydrogen were mixed with an excess of oxygen ($30

Question

$5 	ext{ mL}$ of a gas containing only carbon and hydrogen were mixed with an excess of oxygen ($30 	ext{ mL}$) and the mixture exploded by means of an electric spark. After the explosion, the volume of the mixed gases remaining was $25 	ext{ mL}$. On adding a concentrated solution of potassium hydroxide, the volume further diminished to $15 	ext{ mL}$, the residual gas being pure oxygen. All volumes have been reduced to NTP. What is the molecular formula of the hydrocarbon gas?

Professor PC Thomas Classes · Answer

Volume after explosion = $V_{	ext{CO}_2} + V_{	ext{O}_2(	ext{unreacted})} = 25$ mL. KOH selectively absorbs $	ext{CO}_2$, so $V_{	ext{CO}_2} = 25 - 15 = 10$ mL. The remaining $15$ mL is unreacted $	ext{O}_2$, meaning the $	ext{O}_2$ used = $30 - 15 = 15$ mL. 
Combustion Reaction: $	ext{C}_x	ext{H}_y + (x+y/4)	ext{O}_2 ightarrow x	ext{CO}_2 + (y/2)	ext{H}_2	ext{O}(l)$. 
By comparing volumes (Avogadro's Law): 
$x 	imes 5 = 10 \implies x = 2$. 
$(x + y/4) 	imes 5 = 15 \implies 2 + y/4 = 3 \implies y/4 = 1 \implies y = 4$. 
The hydrocarbon is $	ext{C}_2	ext{H}_4$.