What volume of 0.100M Na3PO4 is required to precipitate all the lead (II) ions from 150.0mL of 0.250M PbNO3)2?

Could someone please explain how to get this.

You need a balanced equation first so you know the molar relationship between the reactants and products.
Pb is +2, PO4 is -3
3Pb(NO3)2(aq) + 2Na3PO4(aq) –> Pb3(PO4)2(s) + 6NaNO3 (aq)
mol Pb = (.250 mol/L Pb(NO3)2 )(.15L) = .0375 mol Pb(NO3)2
.0375 mol Pb(NO3)2 x (2 mol Na3PO4/ 3 mol Pb(NO3)2) = 0.025 mol Na3PO4 needed
0.025 mol Na3PO4 / (0.100 mol/L Na3PO4) = 0.25L = 250 mL 0.100 M Na3PO4

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