How many liters of air is needed for the complete combustion of 4.4 grams 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, divide the available weight of the gas by the weight of 1 mole of this gas.

M C3H8 = 12 x 3 + 8 = 44 grams / mol;

N C3H8 = 4.4 / 44 = 0.1 mol;

The amount of oxygen will be.

N O2 = 0.1 x 5 = 0.5 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 = 0.5 x 22.4 = 11.2 liters;

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

V air = 11.2 / 0.21 = 53.333 liters;