50.0 mL of 0.100 M Na3PO4 is mixed with 150.0 mL of 0.250 M Pb(NO3)2 to produce a solid precipitate of lead (II) phosphate. What mass of this precipitate will be produced?

Question

3Pb(NO3)2 (aq) + 2Na3PO4 (aq)  Pb3(PO4)2 (s) + 6NaNO3 (aq)

Answer 1

3Pb(NO₃)₂ (aq) + 2Na₃PO₄ (aq) ==> Pb₃(PO₄)₂ (s) + 6NaNO₃ (aq) ... balanced equation

Find the limiting reactant. One easy way to do this is to divide moles of each reactant by its own coefficient in the balanced equation.

moles Pb(NO₃)₂ = 0.05 L x 0.100 mol/L = 0.005 moles Pb(NO₃)₂ (÷3 ->0.00167)

moles Na₃PO₄ = 0.150 L x 0.250 mol/L = 0.0375 moles Na₃PO₄ (÷2 -> 0.019)

LIMITING REACTANT = Pb(NO₃)₂

Mass Pb₃(PO₄)₂ = 0.005 mol Pb(NO₃)₂ x 1 molPb₃(PO₄)₂ / 3 mol Pb(NO₃)₂ x 812 g/mol = 1.35 g Pb₃(PO₄)₂