When 120 g of lead nitrate solution interacted with a slight excess of potassium iodide solution, 10.5 g of precipitate fell out.

When 120 g of lead nitrate solution interacted with a slight excess of potassium iodide solution, 10.5 g of precipitate fell out. Calculate the mass fraction of lead nitrate in the starting solution.

1) Let’s compose the reaction equation:
2KI + Pb (NO3) 2 = 2KNO3 + PbI2 ↓
2) Calculate the amount of substance and the mass of lead nitrate:
n (PbI2) = m (PbI2) / M (PbI2) = 10.5 / 461 = 0.023 mol
according to the reaction equation
n (Pb (NO3) 2) = n (PbI2) = 0.023 mol;
3) Determine the mass fraction of lead nitrate in the solution:
m (Pb (NO3) 2) = n (Pb (NO3) 2) ∙ M (Pb (NO3) 2) = 0.023 ∙ 331 = 7.613 g.
ω (K2SO3) = (m (Pb (NO3) 2) / m (solution)) ∙ 100% = (7.613 / 120) ∙ 100 = 6.34%
Answer: 6.34%

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