What volume of a 0.169 moL^ -1 Na3PO4 solution is necessary to completely react with 87.0 mL of 0.106 moL^−1 CuCl2?

Question

Consider the following precipitation reaction: 

2Na3PO4(aq)+3CuCl2(aq)→2Na3PO4(aq)+3CuCl2(aq)→ Cu3(PO4)2(s)+6NaCl(aq)

Answer 1

2Na₃PO₄(aq) + 3CuCl₂(aq) ==>  Cu₃(PO₄)₂(s) + 6NaCl(aq) .. balanced equation

moles CuCl₂ present = 87.0 ml x 1 L / 1000 ml x 0.106 mol / L = 0.00922 moles CuCl₂

moles Na₃PO₄ needed = 0.00922 mols CuCl₂ x 2 mol Na₃PO₄ / 3 mols CuCl₂ = 0.00615 mols Na₃PO₄

volume of Na₃PO₄ = 0.00615 mols x 1 L / 0.169 mols = 0.0364 L = 36.4 mls Na₃PO₄ needed