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Answer to Question #312330 in General Chemistry for Ivan Ackerman

Question #312330

For the chemical reaction



2KI+Pb(NO3)2⟶PbI2+2KNO3




how many moles of lead(II) iodide will be produced from 101 g of potassium iodide?


1
Expert's answer
2022-03-19T02:43:38-0400

Moles of potassium iodide= 101/166= 0.608moles

Mole ratio= 2:1

Moles of lead iodide= 0.304



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