Question

how the mass of reactants compares with the mass of products in 2 KI + Pb(NO3)2 --> PbI2 + 2 KNO3

Answer 1

Explanation:

You have done the hard yakka in quoting the balanced chemical equation. Here, the potassium iodide is clearly the limiting reagent. By the stoichiometry of the reaction, there will be half an equiv of lead iodide per equiv of potassium iodide.

Moles of KI = 30.0⋅g/166.0⋅g⋅mol−1 = ??

So resultant mass of PbI2:

So resultant mass of PbI2:(30.0⋅g×461.01⋅g⋅mol−1)×1/2÷166.0⋅g⋅mol−1 = ?? g

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