Calculate the mass of the precipitate that is formed from barium chloride weighing 83.2 g and sodium sulfate.

When sulfuric acid reacts with barium chloride, an insoluble precipitate of barium sulfate is formed. The chemical reaction is described by the following equation:

BaCl2 + Na2SO4 = BaSO4 + 2NaCl;

From one mole of barium chloride, one mole of barium sulfate is formed.

Let’s find the amount of the substance in 83.2 grams of barium chloride.

M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol;

N BaCl2 = 83.2 / 208 = 0.4 mol;

Find the mass of 0.4 mol of barium sulfate.

Its molar mass is:

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

The mass of insoluble salt will be:

m BaSO4 = 233 x 0.4 = 93.2 grams;