# 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% Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.