What error did i make? (stoichiometry)

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Discussion Overview

The discussion revolves around a stoichiometry problem involving the precipitation of silver iodide from silver acetate and calcium iodide solutions. Participants analyze the calculations and reasoning behind the mass of precipitate formed, exploring concepts related to limiting reagents and the nature of iodine in compounds.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions their calculation, noting a discrepancy of 1.16 grams in their answer and seeks clarification on the mass of silver iodide formed.
  • Another participant points out a potential error in the formula used, questioning the existence of AgI2 and clarifying that iodine does not necessarily come in pairs in compounds.
  • Several participants discuss the concept of limiting reagents, indicating that not all silver and iodine will precipitate due to the stoichiometric relationship in the formation of AgI.
  • There is a correction regarding the use of diatomic iodine versus iodide in the calculations, with participants emphasizing the importance of recognizing the difference in context.
  • One participant suggests that the initial approach to the problem was flawed beyond just the diatomic assumption, indicating that there were multiple issues with the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and the correct interpretation of stoichiometry in this context. There is no consensus on the correctness of the initial approach, and multiple competing viewpoints regarding the precipitate formation and limiting reagents are present.

Contextual Notes

Participants highlight the importance of understanding the stoichiometric relationships and the concept of limiting reagents, but the discussion remains unresolved regarding the specific calculations and assumptions made by the original poster.

Lori
So my answer is off by about 1.16 grams and am wondering if i did something wrong?

Question: What mass of silver iodide precipitates when 25.0 ml of 2.30 M silver acetate solution is mixed with 10.0 ml of 2.35 M calcium iodide solution?

Given:
0.025 L of 2.30 M AgC2H3O2 = 0.0575 mols acetate
0.01 L of 2.35 M caI2 = 0.0235 mols Calcium iodide

Want:
? grams of AgI2

So, i converted 0.0575 mols of acetate to silver and got 6.20 grams Ag
by 0.0575 mols acetate * (1mol Ag/1mol Acetate) *107.87 grams/mol Ag = 6.20 grams Ag

For 0.0235 mols CaI2, i got 5.96 I grams by doing (0.0235 grams CaI2) * (2 mols I/ 1 mol Ca) *126.9 gram/mol...

i added 5.96 grams of Iodine and 6.20 grams of silver together and got 12.17 grams AgI2?
Did i miss something
 
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why is it AgI2 and not AgI? I don't know if AgI2 exists...
 
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HAYAO said:
why is it AgI2 and not AgI? I don't know if AgI2 exists...
Oh that was the mistake... I thought iodine comes with 2 since its diatomic. My bad
 
Lori said:
I thought iodine comes with 2 since its diatomic.

Elemental iodine is diatomic, in compounds it doesn't have to come in pairs.
 
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If you have 0.0575 mol Ag and 0.0470 mol I, does all the silver and all the iodine precipitate?
 
mjc123 said:
If you have 0.0575 mol Ag and 0.0470 mol I, does all the silver and all the iodine precipitate?
In reality, no. It's just that the equilibrium constant (precisely, solubility product constant) is really small, so we have very low concentration of silver and iodine ions in the solvent. Since you have excessive Ag, most likely the concentration of Ag ions are higher than the I ions though with I ions being very very low in concentration, qualitatively speaking.
 
Lori said:
Oh that was the mistake... I thought iodine comes with 2 since its diatomic. My bad
I2, when Iodine is among itself, as the element, or the common way to think of the diatomic element.

I-, the anion, as "iodide", when in an ionic compound as an iodide.
 
HAYAO said:
In reality, no. It's just that the equilibrium constant (precisely, solubility product constant) is really small, so we have very low concentration of silver and iodine ions in the solvent. Since you have excessive Ag, most likely the concentration of Ag ions are higher than the I ions though with I ions being very very low in concentration, qualitatively speaking.
Yes, but the point is that AgI contains equimolar amounts of Ag and I, so once you've precipitated 0.047 mol AgI, there's effectively no I left (in reality a very small amount), so you can't precipitate any more AgI. The remaining silver doesn't precipitate. So adding 5.96g I and 6.20g Ag is wrong. Have you heard of the concept of "limiting reagent"?
 
mjc123 said:
Yes, but the point is that AgI contains equimolar amounts of Ag and I, so once you've precipitated 0.047 mol AgI, there's effectively no I left (in reality a very small amount), so you can't precipitate any more AgI. The remaining silver doesn't precipitate. So adding 5.96g I and 6.20g Ag is wrong. Have you heard of the concept of "limiting reagent"?
You just restated what I've said. What's your point?
 
  • #10
Wait. Was my technique right for this problem? Except I should calculate just one iodine instead of using it as a diatonic element?
 
  • #11
Lori said:
Wait. Was my technique right for this problem?

No.

Except I should calculate just one iodine instead of using it as a diatonic element?

It wasn't the only problem with your approach, just the one that was spotted first.
 

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