Question

What volume of air (containing 20% O₂ by volume) will be required to burn completely 10 cm³ of methane and acetylene?

\[\ce{CH4 + 2O2 → CO2 + 2H2O}\]

\[\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\]

Answer 1

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