Calculate the volume of air required to burn 100L of propane.

The ethane 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 = 100 / 22.4 = 4.46 mol; The amount of oxygen will be.

N O2 = 4.46 x 5 = 22.3 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 = 22.3 x 22.4 = 500 liters;

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

V air = 500/0, 21 = 2 381 liters;