Calculate the amount of oxygen required to burn 54g of sulfur.

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

S + O2 = SO2;

One mole of sulfur reacts with one mole of oxygen to form one mole of sulfur dioxide.

Let’s determine the amount of substance in 54 grams of sulfur.

M S = 32 grams / mol;

N S = 54/32 = 1.6875 mol;

Therefore, the reaction consumed 1.6875 mol of oxygen.

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

V O2 = 1.6875 x 22.4 = 37.8 liters;