Question

\(80\,\text{mL}\) of an organic compound is mixed with \(264\,\text{mL}\) of \(O_2\) and ignited. It gives \(224\,\text{mL}\) of gaseous mixture at NTP. After passing through KOH, \(64\,\text{mL}\) of gas remains. The organic compound is:

• A. \( \mathrm{C_2H_4} \)
• B. \( \mathrm{C_2H_2} \)
• C. \( \mathrm{C_4H_{10}} \)
• D. \( \mathrm{C_3H_6} \)

Hint:

For combustion volume problems:
KOH absorbs only \(CO_2\)
Volumes of gases are proportional to moles at same conditions
Use stoichiometry of combustion to find molecular formula

Answer 1

The Correct Option is B

To identify the organic compound, we need to understand the combustion process of hydrocarbons and how the gases are involved in the reaction. Here's a step-by-step analysis: 

1. Combustion of a hydrocarbon in oxygen forms \(CO_2\) and \(H_2O\). The gaseous products are \(CO_2\) and, if any excess oxygen remains, unreacted \(O_2\).
2. The total volume of the gaseous mixture after combustion is \(224\,\text{mL}\).
3. After passing the gaseous mixture through KOH, the volume reduces to \(64\,\text{mL}\). KOH absorbs \(CO_2\), so the remaining 64 mL should be the unreacted \(O_2\).
4. Thus, the volume of \(CO_2\) formed is \(224\,\text{mL} - 64\,\text{mL} = 160\,\text{mL}\).
5. Initial volume of \(O_2\) was \(264\,\text{mL}\), and 64 mL remained. So, \(264\,\text{mL} - 64\,\text{mL} = 200\,\text{mL}\) of \(O_2\) was consumed in the reaction.
6. The reaction for the hydrocarbon can be generalized as: \(C_xH_y + kO_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\).
7. For \(\text{C}_2\text{H}_2\):
   • The reaction is: \(2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O\).
   • This means: 80 mL of \(C_2H_2\) needs 200 mL of \(O_2\) to produce 160 mL of \(CO_2\), fitting the given data.

Other hydrocarbons do not satisfy this oxygen and \(CO_2\) balance based on the stoichiometry and the volume reduction described.

Conclusion: Since the volume calculations align well with the reaction of \(\text{C}_2\text{H}_2\), the correct answer is indeed \(C_2H_2\), or acetylene.