Hydrocarbons Part 2

Concept:Combustion of a hydrocarbon produces $CO_2$ and $H_2O$.At 273 K, water condenses, so the residual gases are only $CO_2$ and excess $O_2$.KOH absorbs $CO_2$, leaving only the unreacted $O_2$.Chemical Equation / Formula:General combustion: $C_xH_y + \left(x + \frac{y}{4}\right) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O$Explanation:Given: Volume of hydrocarbon = 80 mL, initial $O_2$ = 264 mL.After combustion and cooling, total gas volume ($CO_2$ + excess $O_2$) = 224 mL.After KOH treatment, volume decreases to 64 mL (only excess $O_2$ remains).Therefore, volume of $CO_2$ produced = $224 - 64 = 160$ mL.From the equation, $CO_2$ volume = $80x$ mL, so $80x = 160 \Rightarrow x = 2$.Oxygen consumed = initial $O_2$ - excess $O_2$ = $264 - 64 = 200$ mL.Oxygen consumed also equals $80 \left( x + \frac{y}{4} \right)$ mL.Substitute $x=2$: $80 \left( 2 + \frac{y}{4} \right) = 200$$\Rightarrow 2 + \frac{y}{4} = 2.5 \Rightarrow \frac{y}{4} = 0.5 \Rightarrow y = 2$.Thus the hydrocarbon is $C_2H_2$.Shortcut:From $CO_2$ volume: $x = \frac{160}{80}=2$.Oxygen used = $200$ mL, so $80(2 + \frac{y}{4}) = 200$ gives $y=2$.Answer:Option A: $C_2H_2$