Calculate the volume of oxygen required for the oxidation of 3.1 grams of phosphorus.

The oxidation reaction of phosphorus with oxygen is described by the following chemical reaction equation:

2P + 5/2 O2 = P2O5;

2.5 moles of oxygen are reacted with 2 moles of phosphorus. 1 mol of its oxide is synthesized.

Let’s calculate the available chemical amount of phosphorus.

M P = 31 grams / mol;

N P = 3.1 / 31 = 0.1 mol;

This reaction will require 0.1 x 2.5 / 2 = 0.125 mol of oxygen.

Let’s calculate the volume of oxygen. To do this, multiply the amount of the substance and the standard volume of 1 mole of the gaseous substance. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

V O2 = 0.125 x 22.4 = 2.8 liters;