Calculate the volume of oxygen required to react with 2.2 Ba.

Given:

m (Ba) = 2.2 g

V (O2) = x l

Decision

1. Let us write down the equation of Ba oxidation.

2Ва + О2 = 2ВаО

2. According to the reaction equation, 1 mole of O2 interacts with 2 moles of Ba.

M (Ba) = 137 g / mol, since moles are 2, then m (Ba) = 2 mol * 137 g / mol = 274 g.

1 mole of O2 occupies a volume of 22.4 liters (the molar volume of any gas).

3. We make up the proportion:

274 g of Ba reacts with 22.4 l of O2, and 2.2 g of Ba – with chl of O2.

2.2 / 274 = x / 22.4, hence x = 2.2 * 22.4 / 274, that is, x = 0.18 liters.

Answer: 0.18 liters of oxygen will react with 2.2 g of barium.