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; One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.