After passing 1 m3 of air through Ba (oh) 2, 2.64 g of BaCO3 was formed.

Let’s find the amount of the substance of barium carbonate BaCO3.

M (BaCO3) = 197 g / mol.

n (BaCO3) = 2.64 g: 197 mol = 0.013 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

Ва (OH) 2 + СO2 = ВаСO3 + H2O.

According to the reaction equation, there is 1 mol of CO2 for 1 mol of BaCO3. Substances are in quantitative ratios 1: 1.

n (ВаСO3) = n (СO2) = 0.013 mol.

Let’s find the volume of CO2.

V = n Vn, where Vn is the molar volume of gas equal to 22.4 l / mol.

V = 0.013 mol × 22.4 L / mol = 0.3 L.

ȹ = (0.3 l: 1000 l) × 100% = 0.03%.

Answer: 0.03%.