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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?
Expert's answer
Moles of potassium iodide= 101/166= 0.608moles
Mole ratio= 2:1
Moles of lead iodide= 0.304
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