Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).
V(Pb(NO₃)₂) = 50 mL ÷ 1000 mL = 0.05 L, volume of solution.
c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.
n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).
n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.
n(Pb(NO₃)₂) = 0.005 mol.
n(NaI) = c(NaI) · V(NaI).
n(NaI) = 0.1 mol/L · 0.05 L.
n(NaI) = 0.005 mol; amount of substance.
From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.
n(Pb(NO₃)₂) = 0.005 mol ÷ 2.
n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.
n(NaI) = 0.005 mol; number of moles NaI used.
The limiting reagent is Pb(NO₃)₂.
n(PbI₂) = 0.005 mol.
m(PbI₂) = n(PbI₂) · M(PbI₂).
m(PbI₂) = 0.005 mol · 461 g/mol.
m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.