How much air is required to burn 20 liters of propane?

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

C3H8 + 5O2 = 3CO2 + 4H2O;

According to the coefficients of this equation, 5 oxygen molecules are required to oxidize 1 propane molecule. This synthesizes 3 molecules of carbon dioxide.

Let’s calculate the amount of propane available.

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

N C3H8 = 20 / 22.4 = 0.893 mol;

The amount of oxygen will be.

N O2 = 0.893 x 5 = 4.465 mol;

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

Its volume will be: V O2 = 4.465 x 22.4 = 100 liters;

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

V air = 100/0, 21 = 476.19 liters;