To the potassium phosphate solution was added 322 g of a solution with a mass fraction of silver nitrate of 8%.

To the potassium phosphate solution was added 322 g of a solution with a mass fraction of silver nitrate of 8%. Calculate the mass of the precipitate formed.

Given:
m p (AgNO3) = 322 g
w (AgNO3) = 8% = 0.08
To find:
m (Ag3PO4)
Decision:
3AgNO3 + K3PO4 = Ag3PO4 + 3KNO3
m in (AgNO3) = m p * w = 322 g * 0.08 = 25.76 g
n (AgNO3) = m / M = 25.76 g / 170 g / mol = 0.15 mol
n (AgNO3): n (Ag3PO4) = 3: 1
n (Ag3PO4) = 0.15 mol / 3 = 0.05 mol
m (Ag3PO4) = n * M = 0.05 mol * 419 g / mol = 20.95 g
Answer: 20.95 g