Calculate the mass and volume of oxygen required for interaction with 6.2 g

Calculate the mass and volume of oxygen required for interaction with 6.2 g of phosphorus as a result of the reaction, phosphorus oxide is formed 5.

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 6.2 grams.

M P = 31 grams / mol;

N P = 6.2 / 31 = 0.2 mol;

0.2 mol of phosphorus will react with 0.2 x 5/4 = 0.25 mol 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 = 0.25 x 22.4 = 5.6 liters;

The oxygen weight will be:

M O2 = 16 x 2 = 32 grams / mol;

m O2 = 0.25 x 32 = 8 grams;




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