A mixture of sodium sulfate and sodium bromide weighing 71 g was dissolved in water and treated

A mixture of sodium sulfate and sodium bromide weighing 71 g was dissolved in water and treated with an excess of barium hydroxide, resulting in a precipitate weighing 93.2 g. What is the weight of magnesium chloride in the initial mixture?

Find the resulting amount of barium 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 BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;

N BaSO4 = 93.2 / 233 = 0.4 mol;

To get 0.4 mol of barium sulfate, you need to take the same amount of sodium sulfate.

Let’s calculate its mass:

M Na2SO4 = 23 x 2 + 32 + 16 x 4 = 142 grams / mol;

m Na2SO4 = 0.4 x 142 = 56.8 grams;

The rest is the second salt. Its weight will be 71 – 56.8 = 14.2 grams;