How much air is needed to burn 56 liters of butane?

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

2C4H10 + 13O2 = 8CO2 + 10H2O;

According to the coefficients of this equation, 13 oxygen molecules are required to oxidize 2 butane molecules. In this case, 8 molecules of carbon dioxide are synthesized.

Let’s calculate the amount of butane available.

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

N C4H10 = 56 / 22.4 = 2.5 mol;

The amount of oxygen will be.

N O2 = 2.5 x 13/2 = 16.25 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 = 16.25 x 22.4 = 364 liters;

The oxygen content in the air is 21%.

The required air volume will be:

V air = 364 / 0.21 = 1733.3 liters = 1.733 m3;