Determine the mass of the sediment that is formed by the interaction of 80 g of silver nitrate and 50 g

Determine the mass of the sediment that is formed by the interaction of 80 g of silver nitrate and 50 g of hydrochloric acid.

When silver nitrate interacts with hydrogen chloride, an insoluble silver chloride salt is synthesized.

The process is described by the following equation:

AgNO3 + HCl = AgCl + HNO3;

1 mole of silver nitrate reacts with 1 mole of hydrogen chloride. In this case, 1 mol of silver chloride, insoluble in water, is synthesized.

Let’s calculate the chemical quantities of the available substances.

M AgNO3 = 108 + 14 + 16 x 3 = 170 grams / mol;

N AgNO3 = 80/170 = 0.47 mol;

M HCl = 1 + 35.5 = 36.5 grams / mol;

N HCl = 50 / 36.5 = 1.37 mol;

0.47 mol of nitrate will react with 0.47 mol of acid. In this case, 0.47 mol of silver chloride is synthesized.

M AgCl = 108 + 35.5 = 143.5 grams / mol;

The sediment weight will be:

m AgCl = 0.47 x 143.5 = 67.45 grams;