How many grams of precipitate is formed by the interaction of a solution containing 50 g of copper nitrate (2)

How many grams of precipitate is formed by the interaction of a solution containing 50 g of copper nitrate (2) and a solution containing 50 g of sodium phosphate

Given:
m (Cu (NO3) 2) = 50 g
m (Na3PO4) = 50 g

To find:
m (draft) -?

1) Write the reaction equation:
3Cu (NO3) 2 + 2Na3PO4 => Cu3 (PO4) 2 ↓ + 6NaNO3;
2) Calculate the amount of substance Cu (NO3) 2:
n (Cu (NO3) 2) = m / M = 50/188 = 0.3 mol;
3) Calculate the amount of Na3PO4 substance:
n (Na3PO4) = m / M = 50/164 = 0.3 mol;
4) Determine the amount of substance Cu3 (PO4) 2:
n (Cu3 (PO4) 2) = n (Cu (NO3) 2) / 3 = 0.3 / 3 = 0.1 mol;
5) Calculate the mass of Cu3 (PO4) 2:
m (Cu3 (PO4) 2) = n * M = 0.1 * 382 = 38.2 g.

Answer: The mass of Cu3 (PO4) 2 is 38.2 g.