# Determine the mass of the precipitate that forms when an excess of barium chloride solution is added

Determine the mass of the precipitate that forms when an excess of barium chloride solution is added to a solution containing: a) 52.8 g of ammonium sulfate, b) 200 g of ammonium sulfate

When barium chloride interacts with ammonium sulfate, an insoluble salt of barium sulfate is synthesized.

The process is described by the following equation:

BaCl2 + (NH4) 2SO4 = BaSO4 + 2 NH4Cl;

1 mole of barium chloride reacts with 1 mole of ammonium sulfate. In this case, 1 mol of water-insoluble barium sulfate is synthesized.

Let’s calculate the chemical amount of the available substance.

A.

M (NH4) 2SO4 = (14 + 4) x 2 + 32 + 16 x 4 = 132 grams / mol;

N (NH4) 2SO4 = 52.8 / 132 = 0.4 mol;

0.4 mol of barium chloride reacts with 0.4 mol of sulfuric acid. In this case, 0.4 mol of water-insoluble barium sulfate is synthesized.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;

The sediment weight will be:

m BaSO4 = 0.4 x 233 = 93.2 grams;

B.

M (NH4) 2SO4 = (14 + 4) x 2 + 32 + 16 x 4 = 132 grams / mol;

N (NH4) 2SO4 = 200/132 = 1.515 mol;

1.515 mol of barium chloride reacts with 1.515 mol of sulfuric acid. This synthesizes 1.515 mol of water-insoluble barium sulfate.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;

The sediment weight will be:

m BaSO4 = 1.515 x 233 = 353 grams;

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