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?
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