Calculate the mass of oxygen required for the combustion of 0.31g phosphorus.

The combustion of phosphorus in oxygen is described by the following equation:

4P + 5O2 = 2P2O5;

For the combustion of 4 moles of phosphorus, 5 moles of oxygen are needed.

Find the amount of substance in 0.31 grams of phosphorus.

Its molar mass is 31 grams / mol.

The amount of substance is 0.31 / 31 = 0.01 mol.

To burn such an amount, 0.01 x 5/4 = 0.0125 mol of oxygen will be needed.

One mole of ideal gas takes up a volume of 22.4 liters under normal conditions.

Let us find the volume of 0.0125 mol of oxygen:

V = 0.0125 x 22.4 = 0.28 liters.

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