what mass, in g, of Pb₃(PO₄)₂ (FW 811.54 g/mol) precipitate will be formed

Question

If 475 g of Pb(NO₃)₂ (FW 331.21 g/mol) is added to a solution containing 215 g of K₃PO₄ (FW 212.27 g/mol), what mass, in g, of Pb₃(PO₄)₂ (FW 811.54 g/mol) precipitate will be formed? The balanced equation is 3 Pb(NO₃)₂(aq) + 2 K₃PO₄(aq) → Pb₃(PO₄)₂(s) + 6 KNO₃(aq)

Answer 1

3 Pb(NO₃)₂(aq) + 2 K₃PO₄(aq) → Pb₃(PO₄)₂(s) + 6 KNO₃(aq) ... balanced equation

First, find the limiting reactant. An easy way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever result is less represents the limiting reactant.

For Pb(NO₃)₂ we have 475 g x 1 mol / 331.21 g = 1.434 mols (÷3->0.478

For K₃PO₄ we have 215 g x 1 mol / 212.27 g = 1.012 mols (÷2->0.506)

Therefore, Pb(NO₃)₂ is limiting since 0.478 is less than 0.506.

Finally, use the moles of limiting reactant to find the moles of Pb₃(PO₄)₂ formed:

1.434 mol Pb(NO₃)₂ x 1 mol Pb₃(PO₄)₂ / 3mols Pb(NO₃)₂ x 811.54 g / mol = 388 g Pb₃(PO₄)₂ formed