Determine the mass of the precipitate that is formed by the action of an excess of silver nitrate on 350 g of 8.2%

Determine the mass of the precipitate that is formed by the action of an excess of silver nitrate on 350 g of 8.2% sodium phosphate solution.

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

The process is described by the following equation:

3AgNO3 + Na3PO4 = Ag3PO4 + 3NaNO3;

3 mol of silver nitrate reacts with 1 mol of aluminum phosphate. This synthesizes 1 mol of silver phosphate insoluble in water.

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

M Na3PO4 = 23 x 3+ 31 + 16 x 4 = 164 grams / mol;

N Na3PO4 = 350 x 0.082 / 164 = 0.175 mol;

3 x 0.175 = 0.525 mol of silver nitrate reacts with 0.175 mol of aluminum phosphate. In this case, 0.175 mol of silver phosphate insoluble in water is synthesized.

M Ag3PO4 = 108 x 3 + 31 + 16 x 4 = 419 grams / mol;

The sediment weight will be:

m Ag3PO4 = 419 x 0.175 = 73.325 grams;