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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 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.
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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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