The iron content of iron ore can be determined by titration with a standard \(\mathrm{KMnO}_{4}\) solution. The iron ore is dissolved in \(\mathrm{HCl},\) and all the iron is reduced to \(\mathrm{Fe}^{2+}\) ions. This solution is then titrated with \(\mathrm{KMnO}_{4}\), solution, producing \(\mathrm{Fe}^{3+}\) and \(\mathrm{Mn}^{2+}\) ions in acidic solution. If it required 38.37 \(\mathrm{mL}\) of 0.0198 \(\mathrm{M}\) \(\mathrm{KMnO}_{4}\) to titrate a solution made from 0.6128 \(\mathrm{g}\) of iron ore, what is the mass percent of iron in the iron ore?

First, we need to write down the balanced chemical equation for the redox reaction between \(Fe^{2+}\) ions and \(KMnO_4\) in acidic medium: \(5Fe^{2+} + MnO_4^{-} + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O\) This equation is important as it gives us the stoichiometric relationship between \(Fe^{2+}\) ions and \(KMnO_4\) (or \(MnO_4^{-}\)).

From the balanced chemical equation, we can see that 1 mole of \(MnO_4^{-}\) reacts with 5 moles of \(Fe^{2+}\). Now we will use the given volume and molarity of the \(KMnO_4\) solution used in the titration to calculate the moles of \(Fe^{2+}\) ions present. Moles of \(MnO_4^{-}\) used in the titration = Molarity × Volume in liters Moles of \(MnO_4^{-}\) = \(0.0198\,M \times 0.03837\,L\) Moles of \(MnO_4^{-}\) = \(7.597 \times 10^{-4}\,mol\) Now, use the stoichiometry from the balanced equation to calculate the moles of \(Fe^{2+}\) ions: Moles of \(Fe^{2+}\) = 5 × Moles of \(MnO_4^{-}\) Moles of \(Fe^{2+}\) = \(5 \times 7.597 \times 10^{-4}\,mol\) Moles of \(Fe^{2+}\) = \(3.7985 \times 10^{-3}\,mol\)

Now that we have the moles of \(Fe^{2+}\) ions, we can calculate the mass of iron in the sample. Remember that the molar mass of iron is 55.845 g/mol. Mass of iron = moles × molar mass of iron Mass of iron = \((3.7985 \times 10^{-3}\,mol) \times (55.845\,g/mol)\) Mass of iron = \(0.2121\,g\) Finally, we can calculate the mass percent of iron in the iron ore by dividing the mass of the iron by the mass of the iron ore sample (both in grams) and multiplying by 100: Mass percent of iron = \(\frac{Mass \,of\, iron}{Mass \,of\, iron\, ore\, sample} \times 100\) Mass percent of iron = \(\frac{0.2121\,g}{0.6128\,g} \times 100\) Mass percent of iron = 34.61% Thus, the mass percent of iron in the iron ore is 34.61%.