When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?

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.

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