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

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Discussion Overview

The discussion revolves around calculating the percent yield of PbI2 from a chemical reaction involving CsI and Pb(NO3)2. Participants explore the stoichiometry of the reaction, the concept of purity in reagents, and the correct approach to determining the theoretical yield based on the limiting reagent.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant presents a calculation for percent yield but encounters an issue where the calculated yield exceeds 100%, indicating a potential error in their approach.
  • Another participant questions the necessity of dividing the mass of Pb(NO3)2 by its purity percentage, suggesting it may not be the correct method.
  • Some participants clarify that the mass of pure Pb(NO3)2 should be calculated by multiplying the total mass by the purity percentage rather than dividing.
  • There is a suggestion to recalculate the mass of Pb(NO3)2 and CsI using the correct method to find the actual amounts of the pure compounds available for the reaction.

Areas of Agreement / Disagreement

Participants express differing views on the method of calculating the mass of pure Pb(NO3)2 and CsI, with some advocating for multiplication by purity while others initially suggested division. The discussion remains unresolved regarding the initial calculation method and its implications for the percent yield.

Contextual Notes

Participants highlight the importance of correctly interpreting the purity of the reagents and its impact on stoichiometric calculations. There are unresolved steps in the calculations presented, particularly concerning the transition from mass to moles and the determination of the limiting reagent.

Who May Find This Useful

This discussion may be useful for students working on chemistry homework related to stoichiometry, percent yield calculations, and the implications of reagent purity in chemical reactions.

orchidee7
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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% ?!

Please help :( I'm not sure where I went wrong...
 
Last edited:
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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.
 

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