How many grams of sediment is formed by the interaction of 16.4 grams of “sodium phosphate” and 70 grams of “silver nitrate”.

To solve, we compose the equation of the process:

Na3PO4 + 2AgNO3 = 3NaNO3 + Ag3PO4 – ion exchange, a precipitate of silver phosphate is formed;
Let’s make calculations using the formulas of substances:
M (Na3PO4) = 163.6 g / mol;

M (AgNO3) = 169.8 g / mol;

M (Ag3PO4) = 418.3 g / mol.

Let’s determine the amount of starting substances:
Y (Na3PO4) = m / M = 16.4 / 163.6 = 0.1 mol (deficient substance);

Y (AgNO3) = 0.1 mol since the amount of these substances is 1 mol.

Y (AgNO3) = m / M = 70 / 169.8 = 0.4 mol (substance in excess).

Calculations are made for the substance in deficiency.

Find the mass of the product:
m (Ag3PO4) = Y * M = 0.1 * 418.3 = 41.83 g

Answer: The mass of silver phosphate is 41.83 g