Percent Yield

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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:

Answers and Replies

  • #2
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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.
 
  • #3
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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
 
  • #4
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Is there a point to dividing by that percentage?
 
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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…
 
  • #6
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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?
 
  • #7
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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!!!
 
  • #8
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Very good.
 

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