Answer to Question #169374 in General Chemistry for Fabian Rodriguez

Question #169374

For the reaction Pb(NO3)2 + 2KI → PbI2 + 2KNO3, how many moles of lead iodide are produced from 350. g of potassium iodide?


1
Expert's answer
2021-03-08T05:58:44-0500

You

13:17

1) n(KI)= m(KI)/M(KI)= 350/166=2.108 mole

2) n(PI2)=n(KI)/2=2.108/2= 1.05 mole

Answer: 1.05 mole PbI2


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