How much oxygen is required to burn 620 g of phosphorus containing 15% impurities?

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

4P + 5O2 = 2P2O5;

4 phosphorus atoms interact with 5 oxygen molecules. During the reaction, 2 molecules of pentavalent phosphorus oxide are synthesized.

Let’s calculate the chemical amount of a substance contained in phosphorus weighing 620 grams, containing 15% impurities.

M P = 31 grams / mol;

N P = 620 x 0.85 / 31 = 17 mol;

17 moles of phosphorus will react with 17 x 5/4 = 21.25 moles of oxygen.

Let’s calculate its volume.

To do this, multiply the amount of oxygen by the volume of 1 mole of gas.

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

V O2 = 21.25 x 22.4 = 476 liters;