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

..........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.

10.2g/331.208 = 0.0308

(0.25M)(0.25L) = 0.0625

now what? subtract those two #s?

..........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.

0.0308x 331.208 = 10.2 g

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

To find the mass of PbI2 produced in the reaction, we need to determine the limiting reagent first. The limiting reagent is the one that is completely consumed in the reaction and determines the amount of product formed.

Let's calculate the number of moles of Pb(NO3)2 and KI in their respective solutions.

To find the number of moles of Pb(NO3)2:
Using the formula n = m/M, where n is the number of moles, m is the mass, and M is the molar mass of Pb(NO3)2.
m(Pb(NO3)2) = 10.2 g and M(Pb(NO3)2) = 331.21 g/mol.
n(Pb(NO3)2) = 10.2 g / 331.21 g/mol = 0.0308 mol

To find the number of moles of KI:
n(KI) = C(V), where C is the concentration of KI and V is the volume.
C(KI) = 0.250 M and V(KI) = 0.2500 L = 250.0 mL = 0.2500 L.
n(KI) = 0.250 M * 0.2500 L = 0.0625 mol

Now we need to determine the stoichiometry of the balanced equation to find the ratio between Pb(NO3)2 and PbI2. The balanced equation for the reaction between Pb(NO3)2 and KI is:

Pb(NO3)2 + 2KI -> PbI2 + 2KNO3

From the equation, we can see that one mole of Pb(NO3)2 reacts to form one mole of PbI2. Therefore, the number of moles of PbI2 produced will also be 0.0308 mol.

To find the mass of PbI2 produced:
Using the formula m = n * M, where m is the mass, n is the number of moles, and M is the molar mass of PbI2.
M(PbI2) = 461.01 g/mol.
m(PbI2) = 0.0308 mol * 461.01 g/mol ≈ 14.2 g

Therefore, the mass of PbI2 produced is approximately 14.2 grams.

To find the number of moles of the excess reagent (KI), we need to determine the amount of KI that is left unreacted after the reaction has taken place.

Since one mole of Pb(NO3)2 reacts with 2 moles of KI, the theoretical amount of KI required would be 2 * 0.0308 mol = 0.0616 mol.

To find the number of moles of excess KI:
n(excess KI) = n(initial KI) - n(KI used)
n(excess KI) = 0.0625 mol - 0.0616 mol = 0.0009 mol

Therefore, the number of moles of excess KI is 0.0009 mol.

I hope this explanation helps! Let me know if you have any more questions.