Calculate the mass of the precipitate obtained by the interaction of sodium carbonate with

Calculate the mass of the precipitate obtained by the interaction of sodium carbonate with a mass of 53 g and calcium chloride, the mass of a solution of which is 400 g with a mass fraction of CaCl2 20%. The solution is also given to write down.

The reaction between calcium chloride and sodium carbonate is described by the following chemical reaction equation:

CaCl2 + Na2CO3 = CaCO3 + 2NaCl;

From 1 mol of calcium chloride, 1 mol of calcium carbonate is synthesized. This reaction requires 1 mole of sodium carbonate.

Let’s calculate the chemical amount of a substance contained in 53 grams of sodium carbonate.

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

N Na2CO3 = 53/106 = 0.5 mol;

Let’s calculate the chemical amount of a substance contained in 400 grams of a 20% calcium chloride solution.

M CaCl2 = 40 + 35.5 x 2 = 111 grams / mol;

N CaCl2 = 400 x 0.2 / 111 = 0.721 mol;

During the reaction, 0.5 mol of calcium carbonate will be synthesized.

Let’s calculate the weight of calcium carbonate. To do this, we multiply the amount of the substance by its molar weight.

M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol;

m CaCO3 = 0.5 x 100 = 50 grams;