Stoichiometry problem?

Question

I need help on this problems. 10 grams of lead 2 nitrate is dropped into a solution made from .25 moles potassium iodide. Keeping in mind all nitrates are soluable how many grams of the precipitate will be made?
Hint: solve for each reactant # of grams of precipitate first!


What does this all means and how do I do it. Thanks for helping me

Answer 1

You have to find the limiting reagent, or that reactant with the lowest molar amount. The amount is the molar amount created for the products.

I.E. if the limiting reagent is 0.4 moles, the products will each be 0.4 moles.

multiply by the molecular weight to determine the actual amounts formed.

Answer 2

The only possible reaction is

Pb(NO3)2(aq) + 2KI(aq) = PbI2(solid) + 2 KNO3(aq)

PbI2 is insoluble, so yes,this will happen.

Yuo need to convert g Pb(NO3)2 to moles, see how many moles KI it would need to react, and see if you have enough KI. If you do, Pb(NO3)2 is the limiting reagent and the number of moles PbI2 is just equal to the number of moles Pb(NO3)2 you had to start with. If you don't, KI is the limiting reagent and you will get just 0.125 moles PbI2 (ask yourself why 0.125, not 0.25).

And to finish with, multiply number or moles PbI2 by the formula mass.

Principles: to go between mass and moles, use formula mass. To go from one substance to another, you always need moles and the balanced equation.

The hint is confusing. It means: First see how much PbI2 you would get from 10 g Pb(NO3)2 ifyou had enough KI. Then seehowmuch PbI2 you would get from 0.25 moles KI if you had enough Pb(NO3)2. The smaller answer is the right one.

I think that is harder than what I suggested, but it will also get you the right answer (think about why).

Good luck!