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DAT achiever CHEM Q~

ippie

ippie
10+ Year Member
Jun 12, 2006
162
0
NY
  1. Pre-Dental
    What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?

    8H+ + 5Fe2+ + MnO4- à 5Fe3+ + Mn2+ + 4H2O

    A. 5 * (1000 / 25.00) * 0.050 * 55.85 * (1 / 1.120) * 100 %
    B. 5 * (1000 / 25.00) * 0.050 * 55.85 * 1.120 * 100 %
    C. 5 * (0.050 / 1000) * 25.00 * 55.85 * 1.120 * 100 %
    D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %
    E. 5 * (25.00 / 1000) * 0.050 * 55.85 * 1.120 * 100 %


    Answer is D.
    I don't even understand what they are asking.
    Please, explain me. anybody?
    Thank you~~~
     

    KTDMD

    Membership Revoked
    Removed
    10+ Year Member
    Jan 6, 2006
    47
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    1. Pre-Dental
      ippie said:
      What percentage of iron is present in the ore, if a 1.120 g sample requires 25.00 ml of 0.050 M KMnO4 to fully oxidize all of the dissolved Fe2+?

      8H+ + 5Fe2+ + MnO4- à 5Fe3+ + Mn2+ + 4H2O

      A. 5 * (1000 / 25.00) * 0.050 * 55.85 * (1 / 1.120) * 100 %
      B. 5 * (1000 / 25.00) * 0.050 * 55.85 * 1.120 * 100 %
      C. 5 * (0.050 / 1000) * 25.00 * 55.85 * 1.120 * 100 %
      D. 5 * (25.00 / 1000) * 0.050 * 55.85 * (1 / 1.120) * 100 %
      E. 5 * (25.00 / 1000) * 0.050 * 55.85 * 1.120 * 100 %


      Answer is D.
      I don't even understand what they are asking.
      Please, explain me. anybody?
      Thank you~~~

      Let me try putting in the units and see if this will help:

      {[(5 mol Fe2+/1 mol MnO4-) * 25 ml reagent * (0.050 mol MnO4-/1000 ml reagent) * (55.85 g Fe2+/1 mol Fe2+)] over 1.120 g ore sample} * 100%

      (Note: Maximize your browser screen so the above equation stays in one line and cancel out those units before associating those terms such that they will conform to D.)
       

      asckwan

      Member
      10+ Year Member
      5+ Year Member
      Aug 26, 2005
      66
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      Gainesville, Texas
      1. Pre-Dental
        KTDMD gave a great solution, but just in case you are like me, who needs to see things really in pieces, this is kind of how I approached the problem. I broke it complete down into parts then put them together.

        The question is asking for a percent of a whole, so the whole thing weighs 1.12g, and you want to find grams of Fe. So keep in mind you are looking for Fe2+ g first.

        Part 1: So we have .05M of KMnO4 and 25mL of it. So .05M KMnO4 x .025L = # of moles of KMnO4.

        Part 2: We have 5 moles of Fe2+ for every 1 mole of KMnO4, so in order to get the correct amount of Fe2+ we need to multiply the number of moles of KMnO4 by the number of moles of Fe2+. Therefore, we need to multiply by 5.

        Part 3: We need GRAMS of Fe2+, so we should multiply the # of moles of Fe2+ by it's weight, 55.85.

        Part 4: To make it into percent of whole you will take Grams of Fe2+ and divide by Grams of the whole thing = 1.12

        Part 5: put it all together

        (Moles of KMnO4) x (5 moles Fe/1 mole KMnO4) x (Fe2+ grams) / (Sample weight) x 100

        (.05M x .025L) x (5/1) x (55.85g) / 1.12 x 100

        Hope this helps!!
         
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