A solution of barium chloride was added to a solution of sodium sulfate weighing 120 g with w = 20%

| April 11, 2021 | education

A solution of barium chloride was added to a solution of sodium sulfate weighing 120 g with w = 20%. Determine the mass of the sediment.

To solve, we write down the equation:
1. Na2SO4 + BaCl2 = BaSO4 + 2NaCl – ion exchange, obtained barium sulfate in the form of a white precipitate;
2. Calculations:
M (Na2SO4) = 141.8 g / mol.
M (BaSO4) = 233.2 g / mol.
3. Determine the mass, the amount of the original substance:
W = m (practical) / m (theoretical) * 100;
m (Na2SO4) = 120 / 0.20 = 600 g.
Y (Na2SO4) = m / M = 600 / 141.8 = 4.23 mol.
Y (BaSO4) = 4.23 mol since the amount of substances according to the equation is 1 mol.
4. Find the mass of the product:
m (BaSO4) = Y * M = 4.23 * 233.3 = 486.9 g.
Answer: The mass of the barium sulfate precipitate is 486.9 g.



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