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Percentage of Manganese

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Date Posted: 4/6/2008 8:57:58 PM  Status: Live
Percentage of Manganese
Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
The manganese present in a 6.5326 gram sample of steel was oxidized to permanganate ion and treated with 20.00 mL of a 0.1000 M ferrous sulfate solution in an acidic medium. Then, 31.70 mL of a 0.006000 M potassium permanganate solution were required to titrate the excess iron(II) according to the following reaction.

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

Calculate the percentage of manganese (54.938 g/mol) in the steel sample.

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Date Posted: 5/16/2008 9:32:55 AM  Status: Live
Asker's Rating: Helpful   
Response:
  Given that
  The molarity of Fe2+ = 0.10 M
 The volume of Fe2+ = 20.0 ml
 The number of moles of Fe2+ = molarity * volume
                                               = 0.10 M * 0.02 L
                                               = 0.002 mol
 The volume of permanganate = 31.7 ml
  The molarity of permanganate = 0.006 M
  The number of moles of permanganate = 0.006 M *0.0317 L
                                                             = 0.0001902 mol
 The given reaction
  MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O
 shows one mole of permanganate is equivalent to five moles of Fe2+ ions
  0.0001902 moles of  permanganate is equivalent to 5* 0.001902 mol
                                                                               = 0.000951 mol
 
  The number of moles of Fe2+ consumed for Mn present in sample = 0.002 mol - 0.000951
                                                                                                       = 0.001049 mol
  The number of moles of Mn present in sample =  0.001049 mol /5
                                                                        = 0.0002098 mol
  The mass of Mn = 0.0002098 mol * 54.938 g/mol
                            = 0.0115 g
  Percentage of manganese = (mass of Mn / mass of sample)*100
                                         = (0.0115 g /6.5326g)*100
                                         = 0.176 %



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