Determine the mass of the precipitate that forms when 15 g of a 5% barium chloride solution

Determine the mass of the precipitate that forms when 15 g of a 5% barium chloride solution and a 10% sodium sulfate solution are merged.

The interaction of barium chloride with sodium sulfate leads to the formation of an insoluble precipitate of barium sulfate. The course of this reaction is described by the following equation:

BaCl2 + Na2SO4 = BaSO4 + 2NaCl;

Find the available chemical amount of barium chloride. For this purpose, we divide its weight by its molar mass, equal to the sum of the molar weights of the atoms included in the molecule.

M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol; N BaCl2 = 15 x 0.05 / 208 = 0.0036 mol;

Let’s calculate the added chemical amount of sodium sulfate. To do this, we divide its weight by its molar mass, equal to the sum of the molar weights of the atoms included in the molecule.

M Na2SO4 = 23 x 2 + 32 + 16 x 4 = 142 grams / mol; N Na2SO4 = 15 x 0.1 / 142 = 0.01 mol;

Sodium sulfate is taken in excess. During the reaction, 0.0036 mol of barium sulfate will be synthesized.

Let’s calculate its weight:

For this purpose, we multiply the amount of the sediment substance by its molar mass, equal to the sum of the molar weights of the atoms included in the molecule.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 233 x 0.0036 = 0.8388 grams;