Question

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is:

• A. \(C_2H_4\)
• B. \(C_2H_6\)
• C. \(C_2H_2\)
• D. \(C_4H_{10}\)

Hint:

Apply Gay-Lussac's law. Volume ratio of reactants and products follows stoichiometric coefficients.

Answer 1

The Correct Option is C

Step 1: Combustion Analysis:
Hydrocarbon \(C_xH_y\). Volume = 80 mL.
Initial \(O_2 = 264\) mL.
Total Volume after combustion (cooled) = \(CO_2 + O_{2,residual}\) = 224 mL (Water is liquid).
Volume after KOH = 64 mL. KOH absorbs \(CO_2\).
Therefore, \(O_{2,residual} = 64\) mL.
Volume of \(CO_2 = 224 - 64 = 160\) mL.
Volume of \(O_2\) reacted = Initial - Residual = \(264 - 64 = 200\) mL.
Step 2: Stoichiometry:
Reaction: \(C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + y/2H_2O\) 1. Volume \(CO_2 = x \times V_{HC} \implies 160 = x(80) \implies x = 2\).
2. Volume \(O_2\) reacted = \((x + y/4) \times V_{HC} \implies 200 = (2 + y/4)(80)\).
\[ \frac{200}{80} = 2 + \frac{y}{4} \] \[ 2.5 = 2 + \frac{y}{4} \implies 0.5 = \frac{y}{4} \implies y = 2 \] Step 3: Formula:
\(C_2H_2\) Step 4: Final Answer:
% Option (C) \(C_2H_2\)