A chemical engineer determines the mass percent of iron in

Chapter 4, Problem 80P

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A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to \(\mathrm{Fe}^{2+}\) in acid and then titrating the \(\mathrm{Fe}^{2+}\) with \(\mathrm{MnO}_{4}^{-}\). \(A 1.1081-g\) sample was dissolved in acid and then titrated with \(39.32 mL\) of \(0.03190 \mathrm{M} \mathrm{KMnO}_{4}\). The balanced equation is

\(8 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{MnO}_{4}^{\cdot}(\mathrm{aq}) \rightarrow 5 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Mn}^{2+}(\mathrm{aq})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)

Calculate the mass percent of iron in the ore.

Equation Transcription:

Fe2+

MnO4-

0.03190 M KMnO4.

8H+(aq) + 5Fe2+(aq) + MnO4-(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Text Transcription:

Fe^2+

MnO_4-

A 1.1081-g

39.32 mL

0.03190 M KMnO_4.

8H^+(aq) + 5Fe^2+(aq) + MnO_4-(aq) rightarrow 5Fe^3+(aq) + Mn^2+(aq) + 4H_2O(l)