How much air is required to burn 30 liters of ethane

The ethane oxidation reaction is described by the following chemical reaction equation.

2C2H6 + 7O2 = 4CO2 + 6H2O;

According to the coefficients of this equation, 7 oxygen molecules are required to oxidize 2 ethane molecules. In this case, 4 molecules of carbon dioxide are synthesized.

Let’s calculate the amount of ethane available.

To do this, we divide the available gas volume by the volume of 1 mole of ideal gas under normal conditions.

N C2H6 = 30 / 22.4 = 1.339 mol;

The amount of oxygen will be.

N O2 = 1.339 x 7/2 = 4.687 mol;

Let’s calculate the gas volume. To do this, multiply the amount of substance by the volume of 1 mole of gas.

Its volume will be: V O2 = 4.687 x 22.4 = 104.99 liters;

Taking into account the oxygen content in the air of 21%, the required air volume will be:

V air = 104.99 / 0.21 = 500 liters;