Chemistry: Precipitation Reaction

Question

Consider the precipitation reaction from Na₃PO₄(aq) and CuCl₂(aq).



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

What volume in mL of 0.17 M CuCl₂ solution is necessary to completely react with 63 mL of 0.143 M Na₃PO₄? Report and round your answer to the FIRST decimal place.  Do not enter the unit.

Answer 1

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

moles of Na₃PO₄ present = 0.0.063 L X 0.143 mol/L = 0.00900 moles

moles CuCl2 needed = 0.00900 mol Na3P4 x 3 mol CuCl2/2 mol Na3PO4 = 0.0135 mol

Volume CuCl2: (x L)(0.17 mol/L) = 0.0135 mol

x = 0.0795 L = 79.5 mls of CuCl2