Calculate the mass of the precipitate that forms when 50 g of silver nitrate and sodium phosphate solution are poured.

1.3AgNO3 + Na3PO4 = Ag3PO4 + 3NaNO3;

2.find the chemical amount of silver nitrate:

n (AgNO3) = m (AgNO3): M (AgNO3);

M (AgNO3) = 108 + 14 + 3 * 16 = 170 g / mol;

n (AgNO3) = 50: 170 = 0.2941 mol;

3.According to the reaction equation, the amount of precipitate formed is three times less than silver nitrate:

n (Ag3PO4) = n (AgNO3): 3 = 0.2941: 3 = 0.098 mol;

4.find the mass of silver phosphate:

m (Ag3PO4) = n (Ag3PO4) * M (Ag3PO4);

M (Ag3PO4) = 108 * 3 + 31 + 16 * 4 = 419 g / mol;

m (Ag3PO4) = 0.098 * 419 = 41.062 g.

Answer: 41.062 g.