Calculate the volume of oxygen required to oxidize 3.1 g of phosphorus

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

4P + 5O2 = 2P2O5;

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

Determine the amount of substance contained in 3.1 grams of phosphorus:

N P = 3.1 / 31 = 0.1 mol;

The required amount of oxygen will be:

N O2 = 0.1 x 5/4 = 0.125 mol;

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.

Determine the volume of oxygen:

V O2 = 0.125 x 22.4 = 2.8 liters;