Question

80 mL of an organic compound is mixed with 264 mL of O₂ and ignited. It gives 224 mL of gaseous mixture at NTP. After passing through KOH, 64 mL of gas remains. The organic compound is:

• A. C₂H₆
• B. C₂H₂
• C. C₄H₁₀
• D. C₆H₆

Hint:

For combustion reactions, the volume of CO₂ formed is related to the amount of oxygen consumed. After passing through KOH, only CO₂ remains as the product.

Answer 1

The Correct Option is B

To determine the organic compound, we will follow the analysis given in the problem:

1. The combustion reaction involves an organic compound X, which contains only carbon and hydrogen. When it burns with oxygen, the products are carbon dioxide (CO₂) and water (H₂O).
2. According to the problem, 80 mL of the compound is mixed with 264 mL of O₂ and the combustion produces 224 mL of a gaseous mixture.
3. When the gaseous mixture is passed through KOH (which absorbs CO₂), 64 mL of gas remains. This remaining gas is likely to be H₂O in a gaseous state, since it will not be absorbed by KOH.
4. Therefore, the 224 mL of gaseous mixture consists of:
   • CO₂ (absorbed by KOH)
   • Remaining gas (H₂O as steam/gas), which is 64 mL
5. The volume of CO₂ released can be calculated as: \(224\, \text{mL} - 64\, \text{mL} = 160\, \text{mL}\).
6. We need to write the combustion reactions and match the stoichiometry for each option to find out which compound provides the resulting volumes:
   • For C₂H₆: \(2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O\).
   • For C₂H₂: \(2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O\). The logical flow here indicates that 80 mL of C₂H₂ and 264 mL of O₂ can yield approximately 160 mL of CO₂ and 64 mL of H₂O. This precisely matches the problem statement.
   • For C₄H₁₀: \(2C_4H_10 + 13O_2 \rightarrow 8CO_2 + 10H_2O\).
   • For C₆H₆: \(2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O\).
7. Thus, according the balanced reactions that match the volumes accordingly, the organic compound is C₂H₂ (acetylene).

Therefore, the correct answer is C₂H₂.