Problem 73 - Chemistry: Principles and Reactions

Question

Limonite, an ore of iron, is brought into solution in acidic medium and titrated with \(\mathrm{KMnO}_{4}\). The unbalanced equation for the reaction is $$ \mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q) $$ It is found that a \(1.000-\mathrm{g}\) sample of the ore requires \(75.52 \mathrm{~mL}\) of \(0.0205 \mathrm{M}\) \(\mathrm{KMnO}_{4}\). What is the percent of Fe in the sample?

Step-by-Step Solution

Answer

Answer: The percent of Fe in the limonite sample is 69.22%.

Step 1: Balance the chemical equation

To balance the chemical equation, we need 8 moles of H2O on the right side and 5 moles of H+ ions on the left side. The balanced equation becomes: $$ 5 \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{Fe}^{2+}(a q) \longrightarrow 5 \mathrm{H}^{+}(a q)+8 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q) $$

Step 2: Calculate moles of KMnO4 used

To calculate the moles of KMnO4 used in the titration, we start with the given volume and concentration: Moles of KMnO4 = \(\text{volume (L)} \times \text{concentration (M)}\) Moles of KMnO4 = \((75.52\times10^{-3})\, \text{L} \times 0.0205\,\text{M} = 1.5486 \times10^{-3}\,\text{moles}\)

Step 3: Calculate moles of Fe in the sample

Using the balanced equation and stoichiometry, we can find the moles of Fe in the sample: $$ \frac{1.5486\times10^{-3}\,\text{moles KMnO}_4}{1} \times \frac{8\,\text{moles Fe}^2+}{1\,\text{mole KMnO}_4} = 1.239\times10^{-2}\,\text{moles Fe}^2+ $$

Step 4: Calculate the mass and percent of Fe in the sample

Now that we have the moles of Fe, we can convert it to mass using the molar mass of Fe: Mass of Fe = moles of Fe × molar mass of Fe Mass of Fe = \(1.239\times10^{-2}\,\text{moles} \times 55.85\, \frac{\text{g}}{\text{mole}} = 0.6922\,\text{g}\) Now, we can find the percent of Fe in the sample: Percent of Fe = \(\frac{\text{Mass of Fe}}{\text{Mass of limonite sample}}\times100\) Percent of Fe = \(\frac{0.6922\,\text{g}}{1.000\,\text{g}}\times100 = 69.22\%\) Therefore, the percent of Fe in the limonite sample is 69.22%.

Key Concepts

Stoichiometry in Chemistry

Stoichiometry is a section of chemistry that involves using relationships from the balanced chemical equations to calculate the quantities of reactants and products. It's like a recipe for chemistry where you need to know the precise amounts of ingredients you need and how much you'll end up with.