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Molarity/redox reactions

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data
    What is the molarity of a NaCl solution if 18.3mL of the solution reacted with 13.6mL of 0.1M KMnO4 based on the following unbalanced redox reaction in an acidic solution?

    Cl- + MnO4- yields-> Cl2 + Mn2+


    3. The attempt at a solution

    I did the two half reactions
    Cl- -> Cl2
    MnO4- -> Mn2+

    Do I find the full half reactions (ie: the electrons gained/lost H2O, H+) and then do a molarity problem?
     
  2. jcsd
  3. May 6, 2008 #2

    Borek

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    Staff: Mentor

    Yes - this is in fact simple stoichiometric question, just based on the redox reaction. Find full balanced reaction equation first.
     
  4. May 6, 2008 #3
    Ok I got 10Cl- + 16H+ + 2MnO4- -> 5Cl2 + 2Mn2+ + 8H2O

    Next I did 13.6mL KMnO4 x .1M = .00136moles KMnO4

    what i dont understand is how to find moles of NaCl through the half reaction
     
  5. May 6, 2008 #4
    well i think i got it

    based on the half reaction, there are 5 times as many Cl- ions as there are MnO4- ions.
    therefore there are 5 times as many moles of NaCl than there are moles of KMnO4.
    5 x .00136=.0068moles NaCl

    .0068moles NaCl/.0186L NaCl=.372M NaCl
     
  6. May 6, 2008 #5

    Borek

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    Staff: Mentor

    OK
     
  7. May 9, 2008 #6
    I dont understand this. If we have V of NaCl is 18.3 mL, can we find M by using M=mole/V. So the ratio of NaCl and KMnO4 isn't 1:1 ?
     
  8. May 9, 2008 #7

    Borek

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    Staff: Mentor

    Ratio is given by reaction equation. Take a look at stoichiometric coeffcients.
     
  9. May 9, 2008 #8
    Why do we have to do 10Cl + 16H ......?

    Sorry, I just don't understand your step.
     
  10. May 10, 2008 #9

    Borek

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    Staff: Mentor

    Because these coefficients where necessary to balance the equation. Do you know what it means to balance the equation and why the reaction equations needs to be balanced?
     
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