Problem Solution3CaCl2(aq) + 2Na3PO4(aq) ? Ca3(PO4)2(s) + 6 NaCl(aq)

• 0.200 M x 25.0 mL/1000 mL/L = 0.00500 mol CaCl2

• 0.250 M x 50.0 ml/1000 ml/L = 0.0125 mol Na3PO4

• 0.0125mol/0.00500 mol = 2.50; need 2/3 = 0.667, so we have excess Na3PO4

• 0.00500 mol CaCl2 x (1 mol/3 mol) x (312 g/1 mol) =0.520 g Ca3(PO4)2

• (0.00500 mol x 6 mol/3 mol )/0.0750 L = 0.133 M NaCl

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