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anonymous

  • one year ago

Here is a question from my Chemistry lab on Redox Titration:

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  1. anonymous
    • one year ago
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    A mixture consists of FeSO4 and Fe2(SO4)3. A 0.1223g sample of this mixture was dissolved in water and titrated with 0.00420M KMnO4 solution. The titration required 18.85mL of the KMn04 solution. What percent, by mass, of the mixture was FeSO4?

  2. anonymous
    • one year ago
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    The only formulas I have are: 5 (Fe 2+) + (MnO4 -) + 8 (H +) -> 5 (Fe 3+) + (Mn 2+) + 4 (H2O) 5 (H2C2O4) + 2 (MnO4 -) + 16 (H +) -> 10 (CO2) + 2 (Mn 2+) + 8 (H2O)

  3. anonymous
    • one year ago
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    @mathmate maybe you could help... ?

  4. anonymous
    • one year ago
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    @aaronq

  5. aaronq
    • one year ago
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    so you're oxidizing Fe(II) to Fe{III), so Fe2(SO4)3 is irrelevant so find the moles of KMnO4 used using: \(\sf molarity=\dfrac{moles~of~solute}{L~solution}\) After it's a regular stoichiometry problem, so next, (following the first equation), write a ratio of the chemical species of interest (Fe(II) and MnO^4-), and their coefficients from the balanced reaction. Solve for moles of Fe(II), convert it to grams. Then finally find how the percent of FeSO4 in the sample (divide the mass you found by the mass of the mixture, multiply by 100%).

  6. anonymous
    • one year ago
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    thank you so much @aaronq

  7. aaronq
    • one year ago
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    no problem

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