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Given the reaction: Potassium Iodide reacts with Lead (II) nitrate to produce potassium nitrate and lead (II) iodide. If 1.25g. of Potassium iodide reacts with 2.42g of lead (II) nitrate, how many grams of Lead (II) iodide will be produced?

Guest Nov 26, 2018
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2 KI  + Pb(NO3)2  -------> 2 K NO3  + Pb I2      Is the balanced equation......

 

KI mole weight is 39.0983 +126.90 = 165.9983 gm

From the equation    TWO moles of KI will result in ONE mole PbI2

Pb(NO3)2  mole weight = 331.208 gm  ( and  2 KI = 331.9966 gm  (so I will assume this is a stoichiometric rxn))

 

1.25 g of KI represents   1.25/165.9983 = .00753 mole

   One-half of this will be the number of moles of Pb I2 produced

 

PbI2 mole wt = 207.2 + 2(126.9) = 461 gm

 

1/2 * .00753 mole  *  461 gm/mole = 1.7357 gm of PbI2

 

 

 

(edited)

ElectricPavlov  Nov 26, 2018
edited by ElectricPavlov  Nov 26, 2018

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