# What is the mass of the salt that precipitates during the interaction of barium chloride with a solution of sodium

**What is the mass of the salt that precipitates during the interaction of barium chloride with a solution of sodium sulfate weighing 200 g with a mass fraction of salt of 14.25.**

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 added chemical amount of sodium sulfate. To this end, 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 = 200 x 0.1425 / 142 = 0.2 mol;

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

Let’s determine its weight:

To do this, 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 0.2 = 46.6 grams;