How much air (contains 20% oxygen) is required to burn propane if the result is 67.2 liters of carbon dioxide?

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 carbon dioxide.

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

N CO2 = 67.2 / 22.4 = 3 mol;

The amount of oxygen will be.

N O2 = N CO2 x 5/3 = 3 x 5/3 = 5 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 = 5 x 22.4 = 112 liters;

The required volume of air will be: V air = 112 / 0.21 = 533.33 liters;