Question:
The volume of oxygen required for complete combustion of 0.25 mole of methane at S.T.P. is
The volume of oxygen required for complete combustion of 0.25 mole of methane at S.T.P. is
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For complete combustion of hydrocarbons, use the stoichiometric coefficients to determine the amount of oxygen required.
Updated On: Jan 27, 2026
- 5.6 dm\(^3\)
- 7.46 dm\(^3\)
- 22.4 dm\(^3\)
- 11.2 dm\(^3\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the reaction.
The combustion of methane is represented as: \[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \] This equation shows that for each mole of methane, 2 moles of oxygen are required.
Step 2: Calculation.
For 0.25 moles of methane, the required oxygen is: \[ 0.25 \times 2 = 0.5 \, \text{moles of oxygen} \] At S.T.P., 1 mole of gas occupies 22.4 dm\(^3\), so the volume of oxygen required is: \[ 0.5 \times 22.4 = 5.6 \, \text{dm}^3 \]
Step 3: Conclusion.
The correct answer is (A) 5.6 dm\(^3\).
The combustion of methane is represented as: \[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \] This equation shows that for each mole of methane, 2 moles of oxygen are required.
Step 2: Calculation.
For 0.25 moles of methane, the required oxygen is: \[ 0.25 \times 2 = 0.5 \, \text{moles of oxygen} \] At S.T.P., 1 mole of gas occupies 22.4 dm\(^3\), so the volume of oxygen required is: \[ 0.5 \times 22.4 = 5.6 \, \text{dm}^3 \]
Step 3: Conclusion.
The correct answer is (A) 5.6 dm\(^3\).
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