Question

Given the equation: 2 KI + Pb(NO3)2 --> PbI2 + 2 KNO3, how would you calculate themass of PbI2 produced by reacting of 30.0 g KI with excess Pb(No3)2?

Answer 1

`2KI(aq) + Pb(NO_3)_2(aq) rarr PbI_2(s)darr + 2KNO_3(aq)`.

Moles of `mol` `KI -=` `1/2*mol` `PbI_2(s)`

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 `PbI_2`:

`{(30.0*gxx461.01*g*mol^-1)xx1/2}/(166.0*g*mol^-1)` `=` `??` `g`