Find the volume of air required to burn propane with a mass of 560 grams.

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, 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 = 560/44 = 12.727 mol;

The amount of oxygen will be. N O2 = 12.727 x 5 = 63.635 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 = 66.635 x 22.4 = 1425.4 liters;

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

V air = 1425.4 / 0.21 = 6788 liters;