Calculate the mass of the precipitate formed as a result of the reaction of 7.5 barium chloride

Calculate the mass of the precipitate formed as a result of the reaction of 7.5 barium chloride with 8 g of sodium sulfate.

Let’s execute the solution:

In accordance with the condition of the problem, we will write the data:
BaCl2 + Na2SO4 = BaSO4 + 2NaCl – ion exchange, obtained barium sulfate in the sediment;

Let’s make the calculations:
M (BaCl2) = 208.3 g / mol;

M (Na2SO4) = 141.8 g / mol;

M (BaSO4) = 233.3 g / mol.

Let’s determine the amount of starting substances:
Y (BaCl2) = m / M = 7.5 / 208.3 = 0.04 mol (deficient substance);

Y (Na2SO4) = m / M = 8 / 141.8 = 0.06 mol (substance in excess);

Calculations are carried out for the substance in deficiency.

Y (BaSO4) = 0.04 mol since the amount of substances is 1 mol.

Determine the mass of salt:
m (BaSO4) = Y * M = 0.04 * 233.3 = 9.3 g

Answer: the mass of barium sulfate is 9.3 g