Determine the mass of the sediment that forms when table salt interacts with 50 g of a 3% silver nitrate solution.

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

The process is described by the following equation:

AgNO3 + NaCl = AgCl + NaNO3;

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

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

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

N AgNO3 = 50 x 0.03 / 170 = 0.00882 mol;

0.00882 mol of nitrate will react with 0.00882 mol of acid. This synthesizes 0.00882 mol of silver chloride.

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

The sediment weight will be:

m AgCl = 0.00882 x 143.5 = 1.2657 grams;