200 g of a solution of lead nitrate with an excess of potassium iodide precipitated a mass of 9.25 g, calculate the mass fraction

200 g of a solution of lead nitrate with an excess of potassium iodide precipitated a mass of 9.25 g, calculate the mass fraction of lead nitrate in the original solution.

Reaction: Pb (NO) + 2KI => PbI ↓ + 2KNO
Given: m (PbI) = 9.25g m (Pb (NO)) 200g
Given: M (Pb (NO)) = 331 g / mol M (PbI) = 461 g / mol
Find: ω (Pb (NO₃) ₂ -?
Solution: ν (PbI₂) = m (PbI₂) / M (PbI₂) = 9.25 / 461 = 0.02 ν (Pb (NO₃) ₂) = 0.02 mol (follows from the reaction equation) m (Pb (NO₃ ) ₂) = ν (Pb (NO₃) ₂) × M (Pb (NO₃) ₂ = 331 × 0.02 = 6.62 g ω (Pb (NO₃) ₂) = m (Pb (NO₃) ₂ / m ( Pb (NO)) = 6.62 / 200 = 0.03 (3%)
Answer: ω (Pb (NO₃) ₂) = 0.03 (3%)