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Homework Help: What error did i make? (stoichiometry)

  1. Nov 11, 2017 #1
    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
     
  2. jcsd
  3. Nov 11, 2017 #2

    HAYAO

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    why is it AgI2 and not AgI? I don't know if AgI2 exists...
     
  4. Nov 11, 2017 #3
    Oh that was the mistake... I thought iodine comes with 2 since its diatomic. My bad
     
  5. Nov 12, 2017 #4

    Borek

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    Staff: Mentor

    Elemental iodine is diatomic, in compounds it doesn't have to come in pairs.
     
  6. Nov 13, 2017 #5

    mjc123

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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?
     
  7. Nov 13, 2017 #6

    HAYAO

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    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.
     
  8. Nov 13, 2017 #7

    symbolipoint

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    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.
     
  9. Nov 13, 2017 #8

    mjc123

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    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"?
     
  10. Nov 13, 2017 #9

    HAYAO

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    You just restated what I've said. What's your point?
     
  11. Nov 13, 2017 #10
    Wait. Was my technique right for this problem? Except I should calculate just one iodine instead of using it as a diatonic element?
     
  12. Nov 13, 2017 #11

    Borek

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    Staff: Mentor

    No.

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