When 3.2 g of sulfur were burned, 1.6 g of sulfur dioxide SO2 was formed. How much oxygen was consumed in this case?

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

S + O2 = SO2;

From one mole of sulfur, when reacted with one mole of oxygen, one mole of sulfur dioxide is formed.

Let’s determine the amount of substance in 3.2 grams of sulfur and 1.6 grams of sulfur dioxide.

M S = 32 grams / mol;

N S = 3.2 / 32 = 0.1 mol;

M SO2 = 32 + 16 x 2 = 64 grams / mol;

N SO2 = 1.6 / 64 = 0.025 mol;

Consequently, not all of the sulfur was burnt. Burned out 0.025 mol of sulfur. Wasted 0.025 mol of oxygen.

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

V O2 = 0.025 x 22.4 = 0.56 liters;