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%.

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