Question

Calculate the volume of oxygen required to burn completely a mixture of 18.2 cm³ of ethylene and 9.4 cm³ of hydrogen to carbon dioxide.

\[\ce{C2H4 + 3O2 -> 2CO2 + 2H2O}\]

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

Answer 1

1. Combustion of Ethylene (C₂H₄):

Equation: \[\ce{C2H4 + 3O2 -> 2CO2 + 2H2O}\]

By Gay-Lussac's Law,

1 vol. of C₂H₄ requires = 3 vol. of O₂

For 18.2 cm² of C₂H₄:

Volume of O₂ = 18.2 × 3

= 54.6 cm³

2. Combustion of Hydrogen (H₂):

Equation: \[\ce{2H2 + O2 -> 2H2O}\]

By Gay-Lussac's Law,

2 vol. of H₂ requires = 1 vol. of O₂

For 9.4 cm³ of H₂:

Volume of O₂ = `9.4/2`

= 4.7 cm³

Total Oxygen Required = 54.6 + 4.7 = 59.3 cm³