How much oxygen is required to completely burn 20 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 = 20 / 22.4 = 0.893 mol;

The amount of oxygen will be.

N O2 = 0.893 x 7/2 = 3.126 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 = 3.126 x 22.4 = 70.02 liters;