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A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to Fe^2+ in acid and then titrating the Fe^2+ with MnO4^-. A 1.3510 g sample was dissolved in acid and then titrated with 39.32 mL of 0.03190 M of KMnO4. The balanced equation is:
8H^+(aq)+5Fe^2+(aq)+MnO4^-(aq)->5Fe^2+(aq)+Mn^2+(aq)+4H2O(l)
Calculate the mass percent of iron in the ore.

  • college chemistry -

    moles MnO4^- = M x L
    moles Fe = moles MnO4^- x (5 moles Fe/1 mole MnO4^-) = ??
    g Fe = moles Fe x atomic mass Fe.
    %Fe = (mass Fe/mass sample)*100 = ??

    I know this is just a problem but in real life I doubt a chemical engineer did any of this. In the U. S. at least, I think MOST would have taken the sample to the lab and told a chemist, "Here, analyze this and tell me the percent Fe."

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