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

Question

Master

Chemistry

25 December, 14:10

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

+5

Answers (1)

Know the Answer?

Jase French · Answer 1

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.