When draining 200 g of barium chloride with a mass fraction of 8% 90 g of sodium chromate with a mass

When draining 200 g of barium chloride with a mass fraction of 8% 90 g of sodium chromate with a mass fraction of 10% BaCrO4 precipitated. Calculate the mass fraction of barium chloride in the resulting solution.

Determine 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 = 200 x 0.08 / 208 = 0.077 mol;

Find the available chemical amount of sodium chromate. 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 Na2CrO4 = 23 x 2 +52 + 16 x 4 = 162 grams / mol; N Na2CrO4 = 90 x 0.1 / 162 = 0.055 mol;

0.077 – 0.055 = 0.022 mol of barium chloride will remain in the solution.

Its weight puts:

m2 BaCl2 = 0.022 x 208 = 4.576 grams;

The mass of the solution will decrease by the mass of the sediment. Let’s calculate the weight of 0.055 mol of barium chromate.

M BaCrO4 = 137 +52 + 16 x 4 = 253 grams / mol; m BaCrO4 = 253 x 0.055 = 13.915 grams;

The weight of the solution will be equal: equal to:

M solution = 200 + 90 – 13.915 = 276.085 grams;

The mass fraction of barium chloride in the solution will be:

N BaCl2 = 4.576 / 276.085 = 0.01657 = 1.66%; 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.