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????

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.

After calculating that the mass of Pb(NO3)2 is 10.2g, you need to determine the limiting reagent and the excess reagent.

To do this, you need to compare the moles of each reactant.

Moles Pb(NO3)2 = mass/Molar mass = 10.2g/331.208 g/mol = 0.0308 mol

Moles KI = Molarity x Volume = 0.250 M x 0.250 L = 0.0625 mol

From the balanced equation: Pb(NO3)2 + 2KI -> PbI2 + 2KNO3, you can see that 1 mole of Pb(NO3)2 requires 2 moles of KI.

If the moles of KI are higher than the moles of Pb(NO3)2, then KI is in excess and Pb(NO3)2 is the limiting reagent.

In this case, Pb(NO3)2 is the limiting reagent so we will calculate the moles of excess reagent (KI) by subtracting the moles of KI used from the 2 moles required by 1 mole of Pb(NO3)2.

Moles of excess KI = 2*(moles Pb(NO3)2) - moles KI
= 2*(0.0308 mol) - 0.0625 mol
= 0.0616 mol

Finally, to find the mass of PbI2 produced, you can use the moles of PbI2 determined by the moles of Pb(NO3)2.

Moles of PbI2 = Moles of Pb(NO3)2 = 0.0308 mol

Mass of PbI2 = Moles of PbI2 x Molar mass
= 0.0308 mol x 461.013g/mol
= 14.232 g

Therefore, 14.232 grams of PbI2 will be produced and the number of moles of excess reagent (KI) is 0.0616 mol.

After calculating the moles of Pb(NO3)2 using the formula: grams/molar mass = moles, you have determined that the moles of Pb(NO3)2 is 0.0308 mol.

Next, you calculated the moles of KI using the formula: M (concentration) x L (volume) = moles. From the given information, you obtained that moles of KI is 0.0625 mol.

Now, you can determine the limiting reagent. Remember that the balanced equation for the reaction is: Pb(NO3)2 + 2KI -> PbI2 + 2KNO3

From the equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI. Therefore, if you have enough KI, Pb(NO3)2 will be the limiting reagent. In this case, subtract moles of KI from (2 * moles of Pb(NO3)2) to find the moles of KI in excess.

So, subtract 0.0625 mol from (2 * 0.0308 mol) to find that there is 0.0308 mol of KI in excess.

Now, if you don't have enough KI, KI will be the limiting reagent and Pb(NO3)2 will be in excess. You treat it the same way, except that the moles of Pb(NO3)2 used will be 1/2 moles of KI.

Finally, to find the mass of PbI2 produced, you can use the formula: moles of PbI2 x molar mass = grams.