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Question
What volume of air (containing 20% O2 by volume) will be required to burn completely 10 cm3 of methane and acetylene?
\[\ce{CH4 + 2O2 → CO2 + 2H2O}\]
\[\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\]
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Solution
Methane + Oxygen
\[\ce{\underset{10 cm^3}{\underset{1 V}{CH4} + \underset{2 V}{2O2}} -> CO2 + 2H2O}\]
∴ Volume of \[\ce{O2}\] required = 10 × 2 = 20 cm3
∴ Volume of air = 20 × 5 = 100 cm3
∴ 20 cc of \[\ce{O2}\] is contained in 100 cm3 of air.
Acetylene + Oxygen
\[\ce{\underset{10 cm^3}{\underset{2 V}{2C2H2} + \underset{5 V}{5O2}} -> 4CO2 + 2H2O}\]
∴ Volume of \[\ce{O2}\] required = `5/2 xx 10 = 25` cm3
Total O2 needed = 20 + 25
= 45 cm3
Air conrains 20% oxygen,
∴ Air volume = `45/0.20`
= 225 cm3
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