What amount of salt is obtained by the interaction of 33.3 g of calcium chloride and 16.4 g of sodium phosphate?

Given:
m (CaCl2) = 33.3 g
m (Na3PO4) = 16.4 g

To find:
n (salt) -?

1) 3CaCl2 + 2Na3PO4 => Ca3 (PO4) 2 ↓ + 6NaCl;
2) M (CaCl2) = Mr (CaCl2) = Ar (Ca) * N (Ca) + Ar (Cl) * N (Cl) = 40 * 1 + 35.5 * 2 = 111 g / mol;
3) n (CaCl2) = m (CaCl2) / M (CaCl2) = 33.3 / 111 = 0.3 mol;
4) M (Na3PO4) = Mr (Na3PO4) = Ar (Na) * N (Na) + Ar (P) * N (P) + Ar (O) * N (O) = 23 * 3 + 31 * 1 + 16 * 4 = 164 g / mol;
5) n (Na3PO4) = m (Na3PO4) / M (Na3PO4) = 16.4 / 164 = 0.1 mol;
6) n (Ca3 (PO4) 2) = n (Na3PO4) / 2 = 0.1 / 2 = 0.05 mol;
7) n (NaCl) = n (Na3PO4) * 6/2 = 0.1 * 6/2 = 0.3 mol.

Answer: The amount of Ca3 (PO4) 2 substance is 0.05 mol; NaCl – 0.3 mol.