Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq)→ Cu3(PO4)2(s)+6NaCl(aq)?
What volume of a 0.191 M sodium phosphate (Na3PO4) solution is required to fully react with 87.5 mL of a 0.121 M copper(II) chloride (CuCl2) solution in the following precipitation reaction: 2Na3PO4(aq) + 3CuCl2(aq) → Cu3(PO4)2(s) + 6NaCl(aq)?
1 Answers
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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