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Easy CHEM problem

  1. Jul 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the molarity of the solution (salt and any excess reagent after reaction of 250 mL of 0.05 M Ba(OH)2 with 200 mL of 0.08 M HNO3.

    2. Relevant equations



    3. The attempt at a solution

    I am drawing a blank. Here's the balanced equation

    Ba(OH)2 + 2HNO3 ===> 2H2O + Ba(NO3)s

    Ba(OH) .0125 moles

    2HNO3 .016 moles
     
    Last edited: Jul 29, 2008
  2. jcsd
  3. Jul 30, 2008 #2

    Borek

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    Staff: Mentor

    Simple stoichiometry, this is a limiting reagent question. Your reaction equation is almost OK (chack barium nitrate formula, but I suppose that's a typo) and is a correct first step to solution.
     
  4. Jul 30, 2008 #3
    Why can I not edit my own damn post?
     
  5. Jul 30, 2008 #4

    Borek

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    You can for 30 minutes, or something like that.
     
  6. Jul 30, 2008 #5
    Ok, well, Ba(NO3)2, correct?
     
  7. Jul 30, 2008 #6

    Borek

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    Exactly.

    Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O

    Now try the limiting reagent approach.
     
  8. Jul 30, 2008 #7
    I used the number of moles not grams to find the limiting reagent, which is nitric acid. Only .008 moles of barium nitrate will be formed.

    .008 moles/.450 L = .017 M
     
  9. Jul 30, 2008 #8

    Borek

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    0.008 Ba(NO3)2 is a correct intermediate result. But that's not what you were asked about.
     
  10. Jul 30, 2008 #9
  11. Jul 30, 2008 #10

    Borek

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    It is not all. Excess reagent is still present in the solution in unchanged form.
     
  12. Jul 30, 2008 #11
    Ah. I forgot about that.
     
    Last edited: Jul 30, 2008
  13. Jul 30, 2008 #12

    Borek

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    You should list every ion and its concentration separately. EOT for me.
     
  14. Jul 30, 2008 #13
  15. Jul 31, 2008 #14

    chemisttree

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    End Of Thread, I believe. Do you understand the answer?
     
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