Question

Solve the following numerical problem:

Ethane burns in oxygen according to the chemical equation:

\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]

If 80 ml of ethane is burned in 300 ml of oxygen, find the composition of the resultant gaseous mixture when measured at room temperature.

Answer 1

\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]

Molecular weight of ethane = 30 g/mol

Molecular weight of CO₂ = 44 g/mol

Molecular weight of O₂ = 32 g/mol

In reaction, the number of moles of CO₂ formed is twice as that of ethane.

Therefore, the volume of CO₂ will be 2 × 80 = 160 ml.

In the reaction, the number of moles of O₂ used is 3.5 times that of ethane.

Therefore, the volume of O₂ will be 3.5 × 80 = 280 ml.

Unused oxygen is 300 – 280 ml = 20 ml.