How many liters of oxygen is required to burn 6.2 grams of phosphorus.

The synthesis reaction of phosphorus oxide is described by the following chemical reaction equation:

4P + 5O2 = 2P2O5;

4 phosphorus atoms interact with 5 oxygen molecules. This synthesizes 2 molecules of phosphorus oxide.

Let’s calculate the chemical amount of a substance in phosphorus weighing 6.2 grams.

M P = 31 grams / mol;

N P = 6.2 / 31 = 0.2 mol;

To burn such an amount of phosphorus, 0.2 x 5/4 = 0.25 mol of oxygen is required.

Let’s calculate its volume.

1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

The required volume of oxygen will be:

V O2 = 0.25 x 22.4 = 5.6 liters;