You have calculated the moles of Pb(NO3)2 and KI:
moles Pb(NO3)2 = 0.0308
moles KI = 0.0625
Now we need to determine which of the reactants is the limiting reactant. As you can see from the balanced equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KI. So now let's check if you have enough KI by dividing moles KI by 2:
0.0625/2 = 0.03125
Now, compare the moles of Pb(NO3)2 with the moles of KI required to react fully with Pb(NO3)2.
moles Pb(NO3)2: 0.0308
moles KI required to react fully with Pb(NO3)2: 0.03125
Since there are not enough moles of KI to react with the moles of Pb(NO3)2, KI is the limiting reactant. So, the reaction will proceed until all the KI has reacted, and some amount of Pb(NO3)2 will be left unreacted.
To find the amount of PbI2 produced, we can use the stoichiometry of the reaction:
1 mole of KI produces 1/2 mole of PbI2
Since the moles of KI reacting are 0.0625, the moles of PbI2 produced will be half of that, which is:
0.0625/2 = 0.03125 moles
Now, to find the mass of PbI2 produced, multiply the moles by the molar mass:
Mass of PbI2 = (0.03125 moles) * (461.01 g/mol) = 14.41 g
To find the number of moles of the excess reagent (Pb(NO3)2) remaining, subtract the used moles from the initial moles:
Initial moles Pb(NO3)2: 0.0308
Used moles Pb(NO3)2: 0.03125/2 = 0.015625
Excess moles Pb(NO3)2: 0.0308 - 0.015625 = 0.015175 moles
So, 14.41 g of PbI2 will be produced, and 0.015175 moles of Pb(NO3)2 will be left unreacted as the excess reagent.
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 - DrBob222, Monday, December 13, 2010 at 4:30pm
..........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 - Jenni..can you look again i'm still not getting it thx, Monday, December 13, 2010 at 6:37pm
10.2g/331.208 = 0.0308
(0.25M)(0.25L) = 0.0625
now what? subtract those two #s?
chemistry - DrBob222, Monday, December 13, 2010 at 9:12pm
..........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 - Anonymous, Tuesday, December 14, 2010 at 4:00pm
0.0308x 331.208 = 10.2 g
I am sorry but I still don't know what to do after this????
1 answer
1 answer