what volume of 0.200M Na3PO4 is required to precipitate all the lead (II) ions from 160.0ml of 0.150M PbNO3)2?

Question

looking for detail and clear explaination

Answer 1

we have this equation :
2PO4 3- + 3Pb2+ = Pb3(PO4) 2-

the moles of Pb2+ = the moles of Pb(NO3)2 = 0.160 * 0.150 = 0.024
the moles of PO4 3- = the moles of Na3PO4 0.024 * 2 / 3 = 0.016
the Volume of Na3PO4 = 0.016 / 0.2 = 0.08 L = 80 mL