Chemistry
Chemistry, 20.06.2021 14:00, liljett

249 g of potassium iodide, KI, is mixed with 496.5 g of lead(II) nitrate, Pb(NO3)2. The equation of the reaction is represented below.

[Pb(NO3)2 = 331; KI = 166; PbI2 = 461, KNO3 = 101]

Calculate the number of moles of excess reagent left.

Give your answer to three significant figures.

answer
Answers: 3

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249 g of potassium iodide, KI, is mixed with 496.5 g of lead(II) nitrate, Pb(NO3)2. The equation of...

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