How much oxygen is required to burn 48g of sulfur.

The reaction of sulfur with oxygen occurs in accordance with the following chemical reaction equation:

S + O2 = SO2;

For oxidation of 1 mole of sulfur, 1 mole of oxygen is required.

Let’s find the amount of sulfur.

M S = 32 grams / mol;

N S = 48/32 = 1.5 mol;

To burn this amount of sulfur, 1.5 moles of oxygen are required.

Let’s find its volume.

One mole of gas under normal conditions fills a volume of 22.4 liters.

The oxygen volume will be equal to:

V O2 = 1.5 x 22.4 = 2.24 liters;