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
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.