Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq)→ Cu3(PO4)2(s)+6NaCl(aq)?

What volume of 0.191 M Na3PO4 solution is necessary to completely react with 87.5 mL of 0.121 M CuCl2?

1 Answer

  • Moles of CuCl2:

    0.121 moles CuCl2/liter x 87.5 mL x 1liter/1000 mL = 0.01058 moles CuCl2

    0.01058 moles CuCl2 x 2 moles Na3PO4/3moles CuCl2 = 0.00705 moles Na3PO4

    0.00705 moles Na3PO4 x 1 liter Na3PO4/0.191 moles Na3PO4 = 0.0369 liters

    0.0369 liters Na3PO4 x 1000 mL/1liter = 36.9 mL

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