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

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SUMMARY

The discussion focuses on calculating the percent yield of PbI2 from the reaction of CsI and Pb(NO3)2. The actual yield of PbI2 collected was 10.4025 grams, while the theoretical yield was calculated to be 0.4251 grams, leading to an erroneous percent yield of 244%. The error stemmed from incorrectly dividing the mass of Pb(NO3)2 by its purity percentage instead of multiplying, which resulted in an incorrect calculation of the limiting reagent. The correct approach involves multiplying the mass by the purity percentage to find the actual mass of the reactants.

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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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