How much air is required for the complete oxidation of 3.4 g of hydrogen sulfide?

The oxidation reaction of hydrogen sulfide is described by the following chemical reaction equation.

2H2S + 3O2 = 2H2O + 2SO2;

2 moles of hydrogen sulfide reacts with 3 moles of oxygen. In this case, 2 mol of water and 2 mol of sulfur dioxide are synthesized.

Let’s calculate the chemical amount of hydrogen sulfide.

To do this, divide its weight by the weight of 1 mole of hydrogen sulfide.

M H2S = 2 +32 = 34 grams / mol; N H2S = 3.4 / 34 = 0.1 mol;

To oxidize such an amount of hydrogen sulfide, 0.15 mol of oxygen is needed.

Let’s calculate its volume.

To do this, multiply the amount of substance by the volume of 1 mole of gas (22.4 liters).

V O2 = 0.15 x 22.4 = 3.36 liters;

The volume fraction of oxygen in the air is 21%.

The required air volume will be:

V air = 3.36 / 0.21 = 16 liters;