Answer:
0.0125 moles of PbI2 will be formed during this reaction.
Explanation:
Since you've already figured out the amount of moles of KI formed by using the formula of Molarity being M = n/v, you can simply use mole to mole ratios to figure out how much PbI2 forms within the reaction.
So, every two moles of KI will form one mole of PbI2
0.025molKI * 1molPbI2/2molsKI = 0.0125 moles PbI2
I hope this helps!