Question

2 Na3PO4(aq) + 3 Cu(NO3)2(aq) → Cu3(PO4)2(s) + 6 NaNO3(aq) What mass of Cu3(PO4)2 can be formed when 23.7 mL of a 0.790 M solution of Na3PO4 is mixed with 76.3 mL of a 0.730 M solution of Cu(NO3)2?

Answer 1

2 mole Na3PO4 react with  3 mole Cu(NO3)2(aq) to form one mole Cu3(PO4)2.

23.7 mL of a 0.790 M solution of Na3PO4 = 0.0237 L * 0.790 mole / L = 0.0187 mole Na3PO4.

and

76.3 mL of a 0.730 M solution of Cu(NO3)2 = 0.0763 L * 0.730 mole / L = 0.0557 mole.

thus

0.0187 mole Na3PO4 should react with (3 / 2 * 0.0187 ) = 0.02805 mole Cu(NO3)2 but there is 0.0557 mole Cu(NO32.

so

Na3PO4 is the limiting reactant.

mole of  Cu3(PO4)2 formed = 0.0187 / 2 = 0.00935 mole.

and

mass of Cu3(PO4)2 can be formed = 0.00935 mole * 380.58 g/mole = 3.56 g (answer)