When draining 160 g of a solution of barium nitrate with a mass fraction of 10% and a 5-g solution

When draining 160 g of a solution of barium nitrate with a mass fraction of 10% and a 5-g solution of potassium chromate with a mass fraction of 11%, a precipitate fell out. calculate the mass fraction of potassium nitrate in the resulting solution

Let’s calculate the available chemical amount of barium nitrate. 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 Ba (NO3) 2 = 137 + (14 + 16 x 3) x 2 = 261 g / mol;

N Ba (NO3) 2 = 160 x 0.1 / 261 = 0.0613 mol;

Let’s calculate the available chemical amount of potassium chromate. For this purpose, we divide its weight by its molar mass, equal to the sum of the molar weights of the atoms included in the molecule.

M К2CrO4 = 39 x 2 +52 + 16 x 4 = 194 grams / mol;

N К2CrO4 = 5 x 0.11 / 194 = 0.0028 mol;

In the solution, 0.0028 x 2 = 0.0056 mol of potassium nitrate is formed.

Its weight will be equal to:

M KNO3 = 39 + 14 + 16 x 3 = 101 grams / mol;

m KNO3 = 0.0056 x 101 = 0.5656 grams;

The mass of the solution will decrease by the mass of the sediment. Let us determine the weight of 0.0028 mol of barium chromate.

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

m BaCrO4 = 253 x 0.0028 = 0.7084 grams;

The weight of the solution will be:

M solution = 160 +5 – 0.7084 = 164.2916 grams;

The mass fraction of potassium nitrate in the solution will be equal to: equal to:

N KNO3 = 0.7084 / 164.2916 = 0.00431 = 0.431%;