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Question
What volume of oxygen is required to burn completely a mixture of 22.4 dm3 of methane and 11.2 dm3 of hydrogen into carbon dioxide and steam?
\[\ce{CH4 + 2O2 -> CO2 + 2H2O}\]
\[\ce{2H2 + O2 -> 2H2O}\]
Solution
\[\ce{\underset{(1 vol.)}{CH4} + \underset{(2 vol.)}{2O2} -> CO2 + 2H2O}\]
∴ 22.4 dm3 of CH4 require O2 = 2 × 22.4
= 44.8 dm3
\[\ce{\underset{(2 vol.)}{2H2} + \underset{(1 vol.)}{O2} -> \underset{(2 vol.)}{2H2O}}\]
Now, 2 vols. of H2 require O2 = 1 vol
∴ 11.2 dm3 of H2 require O2 = `(11.2 xx 1)/2`
= 5.6 dm3
∴ Total volume of oxygen required = 44.8 + 5.6
= 50.4 dm3
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