BaCl2 + Na2SO4 = BaSO4 + 2NaCl What is the mass of the precipitate that forms at 50 g of a 4

BaCl2 + Na2SO4 = BaSO4 + 2NaCl What is the mass of the precipitate that forms at 50 g of a 4 percent solution of barium chloride with a solution of sodium sulfate.

1. Calculate the mass of barium chloride contained in the solution:

m (BaCl2) = w (BaCl2) * m (solution) = 0.04 * 50 = 2 g;

2. find the chemical amount of barium chloride:

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

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

n (BaCl2) = 2: 208 = 0.0096 mol;

3. Determine the amount of barium sulfate:

n (BaSO4) = n (BaCl2) = 0.0096 mol;

4.Calculate the mass of the sediment:

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

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

m (BaSO4) = 0.0096 * 233 = 2.2368 g.

Answer: 2.2368 g.