Calculate the volume of oxygen required for complete oxidation of 411 g of barium.

The barium oxidation reaction is described by the following chemical reaction equation:

Ba + ½ O2 = BaO;

1 mole of barium reacts with 0.5 mole of oxygen.

Let’s calculate the chemical amount of a substance that is contained in 411 grams of barium.

M Ba = 137 grams / mol;

N Ba = 411/137 = 3 mol;

To burn such an amount of matter, 3/2 = 1.5 mol of oxygen will be required.

Let’s calculate its volume.

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

V O2 = 1.5 x 22.4 = 33.6 liters;