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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?

 

1
Expert's answer
2022-03-04T06:38:01-0500

Pb(NO3)2 + 2 NaI = PbI2 + 2 NaNO3  


moles of lead (II) iodide = 5moles



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