Calculate the mass of oxygen required for the combustion of 0.31g phosphorus.
July 22, 2021 | education
| 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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