Question

what volume of0.100 M NaPO4 is required to precipitate all the lead(II) ions from 150ml of 0.254 M Pb(NO3)2

Answer 1

3Pb(NO3)2 + 2Na3PO4 --> Pb3(PO4)2 + 6NaNO3
mols Pb(NO3)2 = M x L = ?
Convert mols Pb(NO3)2 to mols Na3PO4. That is mols Pb(NO3)2 x (2 mols Na3PO4/3 mols Pb(NO3)2 = ?
Then grams Na3PO4 = mols x molar mass

Answer 2

.25L