The volume of butane is equal to 22.4 liters of combustion to find the mass of the volume of oxygen.

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

2C4H10 + 13O2 = 8CO2 + 10H2O;

According to the coefficients of this equation, 13 oxygen molecules are required to oxidize 2 butane molecules. In this case, 8 molecules of carbon dioxide are synthesized.

Let’s calculate the amount of butane available.

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

N C4H10 = 22.4 / 22.4 = 1 mol;

The amount of oxygen will be.

N O2 = 1 x 13/2 = 6.5 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 = 6.5 x 22.4 = 145.6 liters;