When adding to a solution of barium chloride weighing 50 g with a mass fraction of 10%

When adding to a solution of barium chloride weighing 50 g with a mass fraction of 10% excess sodium sulfate solution precipitation of a mass (g).

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;

Let’s calculate the available chemical amount of barium chloride. 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 BaCl2 = 137 + 35.5 x 2 = 208 grams / mol; N BaCl2 = 5 x 0.1 / 208 = 0.0024 mol;

With this reaction, the same amount of precipitate will be obtained.

Let’s calculate its weight:

For this purpose, we multiply the weight of the precipitate by its molar mass, which is 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 0.0024 = 0.56 grams;