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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?

1
Expert's answer
2022-03-08T03:43:06-0500

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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