Calculate the mass of the precipitate formed by the interaction of two solutions, one of which contains 21.2 g

Calculate the mass of the precipitate formed by the interaction of two solutions, one of which contains 21.2 g of sodium carbonate (Na2CO3), and the second contains calcium chloride (CaCl2) in a stoichiometric ratio.

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 10.6 grams of sodium carbonate.

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

N Na2CO3 = 21.2 / 106 = 0.2 mol;

During the reaction, the same amount 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 = 0.2 x 100 = 20 grams;