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?

\[\ce{CH4 + 2O2 -> CO2 + 2H2O}\]

\[\ce{2H2 + O2 -> 2H2O}\]

Answer 1

\[\ce{\underset{(1 vol.)}{CH4} + \underset{(2 vol.)}{2O2} -> CO2 + 2H2O}\]

∴ 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