Question

10.0mL of 0.30 M sodium phosphate solution reacts with 20.0mL of 0.20 M lead (II) nitrate solution. what mass of precipitate will from ?

Answer 1

No. of mole of PO4^3- : 10/1000 x 0.30 = 0.003
No. of mole of Pb^2+: 20/1000 x 0.20 = 0.004
2(PO4^3-) + 3(Pb2+) -> Pb3(PO4)2
limiting reagent :Pb^2+
No. of mole of Pb3(PO4)2 formed: (0.004/3) = 0.00133
mass of precipitate (i.e. Pb3(PO4)2 ): (0.004/3) x (207x3+31x2+16x4x2) = 1.08 (g) (corr. to 3 sig. fig.)
Hope this is correct and helpful :P