Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
Convert each to mols product. The smaller value will be the mols PbI2 produced.
1.37 mols Pb(NO3)2 x [1 mol PbI2/1 mol Pb(NO3)2] = 1.37 mols PbI2 produced.
2.56 mols KI x (1 mol PbI2/2 mols KI) = 1.28 mols PbI2 produced.
This is a limiting reagent problem. Obviously you may produced ONLY the smaller amount.
if 1.37 moles of Pb(NO3)2 is mixed with 2.56 moles of KI, how many moles of PbI2 are produced?
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