Question

One mole of a hydrocarbon, on combustion in air, produces two moles of \( \text{CO}_2 \), two moles of \( \text{H}_2\text{O} \), and a large amount of heat. This hydrocarbon is:

• A. ethyne
• B. ethene
• C. ethane
• D. methane

Answer 1

The Correct Option is C

Step 1: Understanding the combustion process:
The combustion of a hydrocarbon involves the reaction of the hydrocarbon with oxygen from the air, resulting in the production of carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)), along with the release of a large amount of heat. The general combustion reaction for a hydrocarbon can be represented as:
\[ \text{Hydrocarbon} + O_2 \rightarrow CO_2 + H_2O + \text{heat} \]

Step 2: Analyze the given data:
From the question, we are told that one mole of the hydrocarbon produces two moles of \( \text{CO}_2 \) and two moles of \( \text{H}_2\text{O} \). This suggests a balanced chemical equation for combustion where the stoichiometry involves 2 moles of carbon (from \( \text{CO}_2 \)) and 4 moles of hydrogen (from \( \text{H}_2\text{O} \)). This is typical for a simple alkane.

Step 3: Identifying the correct hydrocarbon:
Based on the stoichiometry (2 moles of \( \text{CO}_2 \) and 2 moles of \( \text{H}_2\text{O} \)), the hydrocarbon in question must be an alkane with the formula \( \text{C}_2\text{H}_6 \), which is known as ethane.

Step 4: The combustion equation for ethane:
The combustion of ethane can be represented as follows: \[ \text{C}_2\text{H}_6 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} + \text{heat} \] This shows that one mole of ethane produces exactly two moles of \( \text{CO}_2 \) and three moles of \( \text{H}_2\text{O} \), which matches the stoichiometric information provided in the question.

Conclusion:
The correct hydrocarbon is ethane (\( \text{C}_2\text{H}_6 \)), which produces two moles of \( \text{CO}_2 \) and two moles of \( \text{H}_2\text{O} \) on combustion. This combustion reaction is highly exothermic, releasing a large amount of heat.