An excess of sodium carbonate solution is added to 416 g of a 10% barium chloride solution.

An excess of sodium carbonate solution is added to 416 g of a 10% barium chloride solution. The formed precipitate was filtered off, and the solution (filtrate) remaining from the filtration was treated with hydrochloric acid until gas evolution completely ceased. At the same time, 438 g of a 50% solution of hydrochloric acid were spent. Determine the mass of sodium carbonate added to the barium chloride solution.

The interaction of barium chloride with sodium carbonate results in the formation of an insoluble precipitate of barium carbonate. The course of this reaction is described by the following equation:

BaCl2 + Na2CO3 = BaCO3 + 2NaCl;

The interaction of barium carbonate with hydrochloric acid is described by the following equation:

BaCO3 + 2HCl = BaCl2 + CO2 + H2O;

Let’s find the consumed amount of hydrochloric acid:

M HCl = 1 + 35.5 = 36.5 grams / mol; N HCl = 438 x 0.5 / 36.5 = 6 mol;

With this amount of acid, it is possible to dissolve only 3 mol of barium carbonate.

Therefore, 3 mol of sodium carbonate was also added to the original solution.

Let’s calculate its mass:

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 Na2CO3 = 23 x 2 + 12 + 16 x 3 = 106 grams / mol; m Na2CO3 = 106 x 3 = 318 grams;