Calculating Percent Yield of PbI2 in 2CsI + Pb(NO3)2 Reaction

• orchidee7
In summary: Now, can you finish the calculation?Very good. Now, can you finish the calculation?Yes, I can. The theoretical yield would be 0.01214 g of PbI2 (using the stoichiometric ratio of 1 mol PbI2 to 2 mol CsI). Therefore, the percent yield would be:(10.4025 g/0.01214 g) x 100% = 85640.03%In summary, the percent yield of the reaction between 22.9841 grams of 75.25% pure Pb(NO3)2 and 51.2354 grams of 81.21% pure CsI, resulting in 10.4025 grams of PbI

Homework Statement

In one experiment, 22.9841 grams of 75.25% pure Pb(NO3)2 (pure Pb(NO3)2 has a molar mass of 331.2 grams) was mixed with 51.2354 grams of 81.21% pure CsI (pure CsI has a molar mass of 259.80992 grams):

2CsI(aq) + Pb(NO3)2(aq) → 2CsNO3(aq) + PbI2(s)

After the reaction, 10.4025 grams of PbI2(s) (molar mass 461.0 grams) were collected. What is the percent yield of the reaction?

[Ans.: 43.21]

Homework Equations

actual yield/theoretical yield x 100% = percent yield

actual yield = 10.4025 g

The Attempt at a Solution

The equation is balanced and I solved to find that Pb(NO3)2 is the limiting reagent but, when I convert it from moles into grams to do the percent yield, i get a number over 100%.

This is my attempt…

22.9841 g Pb(NO3)2/75.25% = 0.3054 g Pb(NO3)2 x 1 mol/331.2 g x 1 mol PbI2/1 mol Pb(NO3)2
= 9.222 x 10-4 PbI2 (limiting reagent)

51.2354 g CsI/81.21% = 0.6309 g CsI x 1 mol/259.80992 g x 1 mol PbI2/2 mol CsI
= 1.214 x 10-3 mol PbI2

9.222 x 10-4 mol PbI2 x 461.0 g PbI2/1 mol PbI2 = 0.4251 g PbI2

10.4025 g/0.4251 g x 100% = 244% ?!

Last edited:
orchidee7 said:
22.9841 g Pb(NO3)2/75.25% = 0.3054 g Pb(NO3)2
What is this?
A general note on style: "run-on equations" that mimic/list calculator key-stroke sequences not only confuse people trying to follow your work, but also, you.

Bystander said:
What is this?
A general note on style: "run-on equations" that mimic/list calculator key-stroke sequences not only confuse people trying to follow your work, but also, you.

Sorry, it is the mass divided by the purity percentage of the compound

22.9841 g Pb(NO3)2 / 75.25% = 0.3054 g Pb(NO3)2

Is there a point to dividing by that percentage?

Bystander said:
Is there a point to dividing by that percentage?

I just assumed it was necessary as the question had given the molar mass for the pure compounds…

The problem statement has told you that ~ 3/4 of the mass is lead nitrate. How are you going to calculate the mass of lead nitrate?

Bystander said:
The problem statement has told you that ~ 3/4 of the mass is lead nitrate. How are you going to calculate the mass of lead nitrate?

I didn't even think of that…

I would multiply the percent purity by the mass rather than dividing.

22.9841 g Pb(NO3)2 x 0.7525 = 17.2955 g Pb(NO3)2

and

51.2354 g CsI x 0.8121 = 41.6082 g CsI

Thank you so much!

Very good.

1. What is the purpose of calculating percent yield in this reaction?

The purpose of calculating percent yield is to determine the efficiency of the reaction and how much of the desired product, in this case PbI2, was actually produced. This information can be used to improve the reaction conditions and increase the yield in future experiments.

2. How do you calculate percent yield in this reaction?

Percent yield is calculated by dividing the actual yield (the amount of PbI2 produced in the reaction) by the theoretical yield (the amount of PbI2 that should have been produced based on the stoichiometry of the reaction) and multiplying by 100%.

3. What factors can affect the percent yield in this reaction?

The percent yield can be affected by a number of factors including impurities in the reactants, incomplete reactions, and side reactions. Other factors such as temperature, pressure, and reaction time can also impact the yield.

4. How can the percent yield be improved in this reaction?

The percent yield can be improved by ensuring that the reactants are pure and in the correct stoichiometric ratio, using appropriate reaction conditions (e.g. temperature, concentration), and conducting the reaction for a sufficient amount of time. Techniques such as recrystallization and filtration can also help improve the yield by removing impurities.

5. What are the potential sources of error when calculating percent yield in this reaction?

Potential sources of error when calculating percent yield include inaccuracies in measuring equipment, incomplete reactions, and human error in recording data or performing calculations. Other factors such as impurities in the reactants and losses during the transfer of materials can also contribute to errors in the final yield calculation.