When 150 g of a solution of lead nitrate interacted with a small excess of a solution of potassium

When 150 g of a solution of lead nitrate interacted with a small excess of a solution of potassium iodide, 10.45 precipitates were formed. Calculate the mass fraction of lead nitrate in the initial solution

Given:
m solution (Pb (NO3) 2) = 150 g
m (sediment) = 10.45 g

To find:
ω (Pb (NO3) 2) -?

1) Pb (NO3) 2 + 2KI => PbI2 ↓ + 2KNO3;
2) n (PbI2) = m / M = 10.45 / 461 = 0.022 mol;
3) n (Pb (NO3) 2) = n (PbI2) = 0.022 mol;
4) m (Pb (NO3) 2) = n * M = 0.022 * 331 = 7.282 g;
5) ω (Pb (NO3) 2) = m * 100% / m solution = 7.282 * 100% / 150 = 4.85%.

Answer: The mass fraction of Pb (NO3) 2 is 4.85%.