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:
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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.
Step 1: Understanding the reaction.
The reaction involves the combustion of the organic compound with oxygen, producing a gaseous mixture. The gas remaining after passing through KOH is the unreacted oxygen. Step 2: Calculation of gas volumes.
- Volume of gas after reaction: \(224 \, \text{mL}\)
- Volume of gas left after passing through KOH (unreacted O₂): \(64 \, \text{mL}\)
- Volume of CO₂ formed = \(224 \, \text{mL} - 64 \, \text{mL} = 160 \, \text{mL}\)
Now, applying stoichiometric relations:
Let the volume of C₂H₂ be \(x \, \text{mL}\), and the volume of O₂ used in the reaction is \(x + \frac{y}{4}\). Using the relation:
\[
\frac{80}{200} = \frac{2}{2 + \frac{y}{4}}
\]
Solving for \(y\), we get that \(y = 2\), hence the organic compound is \(\text{C}_2\text{H}_2\). Step 3: Conclusion.
The correct answer is (2) C₂H₂.
