Determine the volume of air required to burn 58.1 g of phosphorus to phosphorus oxide (V).

The phosphorus oxidation reaction is described by the following chemical equation:

4P + 5O2 = 2P2O5;

Let’s calculate the chemical amount of acid contained in 58.1 grams of phosphorus.

M P = 31 grams / mol;

N P = 58.1 / 31 = 1.874 mol;

To burn this amount of phosphorus, 1.874 x 5/4 = 2.3425 mol of oxygen will be required.

Let’s calculate its volume.

1 mole of ideal gas normally takes on a volume of 22.40 liters.

V O2 = 2.3425 x 22.40 = 52.47 liters;

The volume fraction of oxygen in the air is 20.95%;

The air volume will be:

V air = 52.47 / 0.2095 = 250.5 liters;