# During the decomposition of a sample of barium carbonate, a gas with a volume of 4.48 liters (in terms of standard equivalent)

**During the decomposition of a sample of barium carbonate, a gas with a volume of 4.48 liters (in terms of standard equivalent) was released. The mass of the solid residue was 50 g. After that, 100 ml of water and 200 g of a 20% sodium sulfate solution were added to the residue. Determine the mass fraction of sodium hydroxide in the resulting solution.**

Let us write the equation for the thermal decomposition of barium carbonate:

ВаСО3 → ВаО + СО2 (1)

Thus, the solid residue is a mixture of the formed barium oxide and undecomposed barium carbonate.

When water is added, barium oxide dissolves:

ВаО + Н2О → Ва (ОН) 2 (2)

and the resulting barium hydroxide reacts further with sodium sulfate:

Ва (ОН) 2+ Na2SO4 → ВаSO4 ↓ + 2NaOH (3)

Barium carbonate is insoluble in water, so it does not go into solution.

We calculate the amount of carbon dioxide released during the calcination of barium carbonate:

n (CO2) = 4.48 l / 22.4 l / mol = 0.2 mol,

From equation (1): n (BaO) = n (CO2) = 0.2 mol,

m (ВаО) = n (ВаО) ∙ M (ВаО) = 0.2 mol ∙ 153 g / mol = 30.6 g.

Let us determine which of the reagents Ba (OH) 2 or Na2SO4 fully reacts.

We calculate the mass and amount of sodium sulfate substance:

m (Na2SO4) in – va = m (Na2SO4) p – pa ∙ ω (Na2SO4) = 200 g ∙ 0.2 = 40 g

n (Na2SO4) = m (Na2SO4) in-va / M (Na2SO4) = 40g / 142g / mol = 0.282 mol.

From equation (2): n (BaO) = n (Ba (OH) 2) = 0.2 mol.

This means that sodium sulfate is taken in excess, and barium hydroxide reacts completely.

We calculate the amount of substance and the mass of the formed sodium hydroxide:

From equation (3): n (NaOH) = 2 ∙ n (Ba (OH) 2) = 2 ∙ 0.2 mol = 0.4 mol

m (NaOH) in islands = n (NaOH) ∙ M (NaOH) = 0.4 mol ∙ 40 g / mol = 16 g.

Let’s calculate the mass of the resulting solution:

mcon.p-pa = m (BaO) + m (H2O) + m (Na2SO4) solution – m (BaSO4)

m (H2O) = ρ (H2O) ∙ V (H2O) = 1 g / ml ∙ 100 ml = 100 g

From equation (3): n (BaSO4) = n (Ba (OH) 2) = 0.2 mol

m (ВаSO4) = n (ВаSO4) ∙ M (ВаSO4) = 0.2 g / mol ∙ 233 mol = 46.6 g.

mcon.r-pa = m (BaO) + m (H2O) + m (Na2SO4) solution – m (BaSO4) = 30.6 g + 100 g + 200 g – 46.6 g = 284g.

Mass fraction of sodium hydroxide in solution is equal to:

ω (NaOH) = m (NaOH) / micron.r-pa = 16 g / 284 g = 0.0563 (5.63%).

Answer: ω (NaOH) = 5.63%.