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Answer to Question #305505 in General Chemistry for Mcquay
Question #305505
Pb(NO3)2 + 2 NaI = PbI2 + 2 NaNO3
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
1 mol Pb(NO3)2 - 1 mol PbI2
5 mol Pb(NO3)2 - x mol PbI2
x = 5 mol
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