What error did i make? (stoichiometry)

[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

 

[HAYAO]

why is it AgI₂ and not AgI? I don't know if AgI₂ exists...

 

[Lori]

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

 

[Borek]

I thought iodine comes with 2 since its diatomic.

Elemental iodine is diatomic, in compounds it doesn't have to come in pairs.

 

[mjc123]

If you have 0.0575 mol Ag and 0.0470 mol I, does all the silver and all the iodine precipitate?

 

[HAYAO]

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.

 

[symbolipoint]

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.

 

[mjc123]

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"?

 

[HAYAO]

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?

 

[Lori]

Wait. Was my technique right for this problem? Except I should calculate just one iodine instead of using it as a diatonic element?

 

[Borek]

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.