# What error did i make? (stoichiometry)

• Lori
Other issues include not considering the limiting reagent and not balancing the chemical equation.In summary, the conversation involves a discussion about calculating the mass of silver iodide precipitate when mixing silver acetate and calcium iodide solutions. The error in the calculation was due to using iodine as a diatomic element instead of considering it as an anion in the compound. Other issues mentioned were not considering the limiting reagent and not balancing the chemical equation.
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

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

Lori
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.

Lori
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?

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

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.

## 1. What is stoichiometry?

Stoichiometry is the branch of chemistry that deals with the quantitative relationships and calculations of the amounts of reactants and products involved in a chemical reaction.

## 2. What is the purpose of stoichiometry?

The purpose of stoichiometry is to determine the exact amounts of reactants needed to produce a desired amount of product, or to calculate the theoretical yield of a reaction based on the given amounts of reactants.

## 3. What is the most common error in stoichiometry calculations?

The most common error in stoichiometry calculations is using incorrect or inaccurate molar ratios. It is important to double check the coefficients in a balanced chemical equation to ensure the correct ratios are being used.

## 4. How can I avoid making errors in stoichiometry?

To avoid making errors in stoichiometry, it is important to double check all calculations and to use accurate and precise measurements. It is also helpful to use dimensional analysis and to understand the concept of molar ratios.

## 5. What should I do if I realize I made an error in my stoichiometry calculation?

If you realize you made an error in your stoichiometry calculation, you should reevaluate your work and try to identify where the error occurred. It may be helpful to double check your calculations and molar ratios. If necessary, redo the calculation using the correct values.

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