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:

Let the hydrocarbon be \(C_xH_y\).
Volume of hydrocarbon taken = 80 mL.
Initial volume of oxygen = 264 mL.

After combustion and cooling, the total gaseous volume consists of \(CO_2\) and unreacted \(O_2\) (water condenses as liquid).
Total volume after combustion = 224 mL.

After passing the gases through KOH solution, the remaining volume is 64 mL.
Since KOH absorbs \(CO_2\), the remaining gas is unreacted oxygen.

Residual \(O_2 = 64\) mL.
Volume of \(CO_2 = 224 - 64 = 160\) mL.
Volume of \(O_2\) consumed = \(264 - 64 = 200\) mL.

Step 2: Stoichiometry:

The general combustion reaction is:
\[ C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O \] From volume relationships:

1. Volume of \(CO_2\):
\[ x \times 80 = 160 \] \[ x = 2 \] 2. Volume of \(O_2\) consumed:
\[ (2 + \frac{y}{4}) \times 80 = 200 \] \[ 2 + \frac{y}{4} = 2.5 \] \[ \frac{y}{4} = 0.5 \] \[ y = 2 \]
Step 3: Molecular Formula:

The hydrocarbon is \(C_2H_2\).

Step 4: Final Answer:

The correct option is (C) \(C_2H_2\).