Hello, Ra,
Assuming the equation is balanced (it is), it tells us that we need twice as many moles of KI as Pb(NO3)2.
2KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2KNO3(aq)
[2 moles KI/(1 mole Pb(NO3)2) = so the ratio is 2:1]
Calculate the actual number of moles of each. The definition of "M" (Molar) is moles/liter. We can calculate the number of moles by multiplying the concentration (M) times the volume, as long as it is also expressed in liters. We are given the volume in ml, so to get liters, divide by 1000. (1000ml/Liter). [Use 3 sig figs since all the original data has 3 significant figures].
Moles of each is then calculated:
- KI: 20ml *(1 Liter/1000 ml) * (0.5 moles/Liter). Liters cancel, ml cancels, and we're left with moles.
= 0.0375 moles KI
- Pb(NO3)2: 50ml*(1 Liter/1000ml)*(1 mole/Liter) = 0.50 moles Pb(NO3)2
Since we need twice as many moles of KI as Pb(NO3)2, we can see that if all of the KI were consumed, it would only need 0.0188 moles of Pb(NO3)2, We have far more than that (0.50 moles of Pb(NO3)2), so that makes KI the limiting reagent.
The sentence ". . . determine which reactant is in excess the mass of lead(II) iodide that forms." may be missing the word "and." I'll rephrase it:
. . .determine which reactant is in excess AND the mass of lead(II) iodide that forms.
We know that the reaction is limited to the amount of KI, so we'll use that to calculate the mass of PbI2 formed. The equation states that we get one mole of PbI2 for every 2 moles of KI consumed. If all of the KI reacts, we would get:
(0.0188 moles KI)*(1 mole PbI2/2 moles KI) = 0.00938 moles PbI2
Hardly worth the effort, IMHO.
I hope this helps,
Bob
