Hey guys so I was absent for my class and my teacher STILL wants me to figure out how to do limiting reactant and theoretical yield I tried Khan academy and it was no use. So basically the first question is :
My data is : Mass of lead nitrate = 5.03g and Mass of Potassium Iodide is 8.09g. And the mass of my yellow precipitate is 6.66g. And the balanced equation is 2KI + Pb(NO3)2 --> PbI2 + 2K(NO)3
I was able to get this far but now im stuck and here are the questions:
- Find limiting reactant
- Find theoretical yield
- Find actual yield.
You were able to get "how far"? I don't see anything but the question and the numbers. I can work the problem for you which won't help you. If you've watched Kahn academy, what do you not understand now. Let's take it from there. How far can you get on your own and where is it confusing. The 6.66 g is the actual yield.
To find the limiting reactant, you need to compare the amounts of each reactant and determine which one will run out first, limiting the amount of product formed.
Let's start by calculating the moles of each reactant using their respective molar masses.
1. Calculate the moles of lead nitrate (Pb(NO3)2):
Moles = Mass / Molar Mass
Moles of Pb(NO3)2 = 5.03g / (207.2 g/mol + 3 * (14.01 g/mol + 16.00 g/mol))
= 5.03g / 331.22 g/mol
≈ 0.0152 mol
2. Calculate the moles of potassium iodide (KI):
Moles = Mass / Molar Mass
Moles of KI = 8.09g / (39.10 g/mol + 1.01 g/mol)
= 8.09g / 40.11 g/mol
≈ 0.202 mol
Now, let's determine the limiting reactant by comparing the ratio of moles of reactants to the ratio in which they are needed according to the balanced equation:
3. Calculate the moles ratio based on the balanced equation:
2 mol KI : 1 mol Pb(NO3)2
This means that for every 2 moles of KI, we need 1 mole of Pb(NO3)2.
4. Compare the calculated moles ratio with the actual moles ratio:
Moles ratio of KI / Moles ratio of Pb(NO3)2 = 0.202 mol / 0.0152 mol
≈ 13.29
Since the moles ratio is significantly higher than the stoichiometric ratio (2:1), it means that there is an excess of KI. Pb(NO3)2 is the limiting reactant because it will be completely consumed before KI.
Next, let's find the theoretical yield, which is the maximum amount of product that can be formed based on the limiting reactant:
5. Calculate the theoretical moles of yellow precipitate (PbI2) that can be formed:
Using the balanced equation, we can see that 1 mole of Pb(NO3)2 produces 1 mole of PbI2.
Therefore, the moles of PbI2 = moles of Pb(NO3)2 = 0.0152 mol
6. Calculate the mass of the theoretical yield of PbI2:
Mass = Moles × Molar Mass
Mass of PbI2 = 0.0152 mol × (207.2 g/mol + 2 * (126.90 g/mol))
= 0.0152 mol × 461.00 g/mol
≈ 7.00 g
Finally, to find the actual yield, you compare the mass of the yellow precipitate (6.66g) that you obtained in the experiment with the theoretical yield (7.00g) calculated above:
7. Calculate the percent yield:
% Yield = (Actual Yield / Theoretical Yield) × 100
% Yield = (6.66g / 7.00g) × 100
≈ 95.14%
So, the limiting reactant in this reaction is Pb(NO3)2, the theoretical yield of PbI2 is approximately 7.00 grams, and the actual yield is 6.66 grams with a percent yield of approximately 95.14%.