# chemistry

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10.2 g of Pb(NO3)2 are dissolved in 100 mL of water, and added to 250.0 mL of a 0.250 M solution of KI. Determine the mass in grams of Pbl2 produced, and find the number of moles of the excess reagent. PLZ explain

• chemistry - ,

..........Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
moles Pb(NO3)2 = grams/molar mass = ??
moles KI = M x L = ??

1 mol Pb(NO3)2 will produce 1 mole PbI2 and use 2 moles KI. DO YOU HAVE THAT MUCH KI? If so, then Pb(NO3)2 is the limiting reagent and KI is the excess reagent. Simply subtract moles KI from (2*moles Pb(NO3)2 to find moles KI in excess.
If you DON'T have enough KI, then KI is the limiting reagent and Pb(NO3)2 will be excess. You treat it the same way EXCEPT that moles Pb(NO3)2 used will be 1/2 moles KI.
Post your work if you get stuck.

• chemistry - ,

10.2g/331.208 = 0.0308

(0.25M)(0.25L) = 0.0625

now what? subtract those two #s?

• chemistry - ,

..........Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
initial...0.0308....0.0625.....0......0
change...-0.0308...0.0616.+0.0308.+0.0308
final.....0.......0.00090...0.0308....etc

First, the change line.
From the equation you can see that 1 mole Pb(NO3)2 will require 2 moles KI; therefore, I multiplied moles Pb(NO3)2 by 2 to arrive at moles KI used. Why didn't I do it the other way around (and it is possible to have done that but you know you started with the wrong one once that happens). Suppose we DID start with 0.0625 moles KI, how many moles Pb(NO3)2 will react with ALL of the KI. That is 0.0625 x (1 mole Pb(NO3)2/2 moles KI) = 0.0625*(1/2) = 0.03125 AND WE DON'T HAVE THAT MUCH Pb(NO3)2 so lead nitrate is the limiting reagent.
final line. Yes, we subtract the two moles to determine which one remains and how much is unreacted BUT we must first determine the limiting reagent as I did in the change line.
To convert moles PbI2 to grams, conert moles PbI2 x molar mass = grams.

• chemistry - ,

0.0308x 331.208 = 10.2 g

I am sorry but I still don't know what to do after this????