# When 40 g of a 20.8-solution of barium chloride react with a solution of sulfuric acid

September 29, 2021 | education

| **When 40 g of a 20.8-solution of barium chloride react with a solution of sulfuric acid, a precipitate of barium sulfate with a mass of?**

1. Let’s compose the equation of the proceeding reaction:

BaCl2 + H2SO4 = BaSO4 + 2HCl;

2. Calculate the mass of barium chloride:

m (BaCl2) = w (BaCl2) * m (solution BaCl2) = 0.208 * 40 = 8.32 g;

3. Let’s calculate the chemical amount of barium chloride:

n (BaCl2) = m (BaCl2): M (BaCl2);

M (BaCl2) = 137 + 35.5 * 2 = 208 g / mol;

n (BaCl2) = 8.32: 208 = 0.04 mol;

4. Determine the amount of barium sulfate:

n (BaSO4) = n (BaCl2) = 0.04 mol;

5. Set the sediment mass:

m (BaSO4) = n (BaSO4) * M (BaSO4);

M (BaSO4) = 137 + 32 + 4 * 16 = 233 g / mol;

m (BaSO4) = 0.04 * 233 = 9.32 g.

Answer: 9.32 g.

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