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60 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 m lead(II) nitrate. How many grams of lead (II) iodide will precipitate?
5 years ago

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Answered by oobleck
2KI + Pb(NO3)2 = PbI2 + 2KNO3
You will get half as many moles of PbI2 as you use moles of KI
so now just see which gets used up first, and that will tell you how many grams are produced.
5 years ago

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