Calculate the mass of the precipitate that precipitates when 500 g of 2% potassium iodide solution is poured
| November 25, 2020 | education
Calculate the mass of the precipitate that precipitates when 500 g of 2% potassium iodide solution is poured with an excess of lead nitrate solution.
m (KI) = (2% * 500) / 100 = 10g
2KI + Pb (NO3) 2 = PbI2 + 2 KNO3
v (KI) = 10g / 166g / mol = 0.06 mol
v (PbI2) = 0.06 / 2 = 0.03 mol
m (PbI2) = 0.03 * 461 g / mol = 13.83 g – sediment mass
Answer: 13.83 g
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