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Answer to Question #305482 in General Chemistry for Mcquay
Question #305482
Pb(NO3)2 + 2 NaI = PbI2 + 2 NaNO3
2) How many moles of lead (II) iodide were produced when 5 moles of lead (II)nitrate were used in the reaction?
Expert's answer
Pb(NO3)2 + 2 NaI = PbI2 + 2 NaNO3
moles of lead (II) iodide = 5moles
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