# During the decomposition of a sample of barium carbonate, a gas with a volume of 1.12 liters (in terms of standard units)

**During the decomposition of a sample of barium carbonate, a gas with a volume of 1.12 liters (in terms of standard units) was released. The mass of the solid residue was 27.35 g. After that, 73 g of a 30% hydrochloric acid solution were added to the residue. Determine the mass fraction of hydrochloric acid in the resulting solution.**

When barium carbonate decomposes, barium oxide is formed and carbon dioxide is released:

ВаСО3 → t ВаО + СО2

Let’s calculate the amount of carbon dioxide released during the calcination of barium carbonate:

n (CO2) = 1.12 l / 22.4 l / mol = 0.05 mol,

therefore, as a result of the decomposition reaction of barium carbonate, 0.05 mol of barium oxide was formed, and 0.05 mol of barium carbonate also reacted. Let us calculate the mass of the formed barium oxide:

m (ВаО) = 153 g / mol ∙ 0.05 mol = 7.65 g.

We calculate the mass and amount of the substance of the remaining barium carbonate:

m (BaCO3) rest. = 27.35 g – 7.65 g = 19.7 g

n (BaCO3) rest = 19.7 g / 197 g / mol = 0.1 mol.

Both components of the solid residue react with hydrochloric acid – the formed barium oxide and the remaining barium carbonate:

ВаО + 2HCl → ВаСl2 + Н2O

ВаСО3 + 2HCl → ВаСl2 + CO2 ↑ + Н2O.

We calculate the amount of matter and the mass of hydrogen chloride interacting with barium oxide and carbonate:

n (HCl) = (0.05 mol + 0.1 mol) ∙ 2 = 0.3 mol;

m (НСl) = 36.5 g / mol ∙ 0.3 mol = 10.95 g.

Let’s calculate the mass of the remaining hydrogen chloride:

m (HCl) rest. = 73 g ∙ 0.3 – 10.95 g = 10.95 g

We calculate the mass-final solution:

mcon.r-ra = m (НCl) solution –m (СО2) = 27.35 g + 73g– 4.4 g = 95.95 g.

The mass fraction of the remaining hydrochloric acid in the solution is equal to:

ω (HCl) = m (HCl) rest. / mcon.r-pa = 10.95 g / 95.95 g = 0.114 (11.4%).

Answer: ω (HCl) = 11.4%.