An excess of copper (II) chloride solution was added to 106 g of a solution with a mass fraction of potassium

An excess of copper (II) chloride solution was added to 106 g of a solution with a mass fraction of potassium phosphate of 4%. Determine the mass of the precipitated sediment.

1. Let us write the equation for the reaction of solutions of copper chloride (2) with potassium phosphate:

3CuCl2 + 2K3PO4 = Cu3 (PO4) 2 ↓ + 6KCl;

2.Calculate the mass of potassium phosphate:

m (K3PO4) = w (K3PO4) * m (solution) = 0.04 * 106 = 4.24 g;

3.Calculate the chemical amount of potassium phosphate:

n (K3PO4) = m (K3PO4): M (K3PO4);

M (K3PO4) = 3 * 39 + 31 + 4 * 16 = 212 g / mol;

n (K3PO4) = 4.24: 212 = 0.02 mol;

4.determine the amount of sediment:

n (Cu3 (PO4) 2) = n (K3PO4): 2 = 0.02: 2 = 0.01 mol;

5. find the mass of copper phosphate:

m (Cu3 (PO4) 2) = n (Cu3 (PO4) 2) * M (Cu3 (PO4) 2);

M (Cu3 (PO4) 2) = 3 * 64 + 2 * 31 + 8 * 16 = 382 g / mol;

m (Cu3 (PO4) 2) = 0.01 * 382 = 3.82 g.

Answer: 3.82 g.