Question

A 0.2 g sample of pyrolusite is analyzed for manganese content as follows. Add 50.0 mL of 0.1 M solution of ferrous ammonium sulfate to reduce the MnO2 to Mn2 . After reduction is complete, the excess ferrous ion is titrated in acid solution with 0.02 M KMnO4, requiring 15.0 mL. Calculate the percent manganese in the sample as Mn3O4.

Answer 1

Answer:

66.7%

Explanation:

The reaction for the titration of the excess ferrous ion is:

• 5Fe⁺² + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

We calculate the moles of Fe⁺² from the used moles of KMnO₄:

• 0.02 M * 15.0 mL = 0.30 mmol KMnO₄

• 0.3 mmol KMnO₄ * $\frac{5mmolFe^{+2}}{1mmolKMnO_4}$ = 1.5 mmol Fe⁺²

Then we substract those 0.30 mmol from the original amount used:

• 0.1 M * 50.0 mL = 5.0 mmol Fe⁺²

• 5.0 - 1.5 = 3.5 mmol Fe⁺²

The reaction between ferrous ammonium sulfate and MnO₂ is:

• 2Fe⁺² + MnO₂ + 4H⁺ → 2Fe³⁺ + Mn²⁺ + 2H₂O

So we convert those 3.5 mmol Fe⁺² that were used in this reaction to MnO₂ moles:

• 3.5 mmol Fe⁺²  * $\frac{1mmolMnO_2}{2mmolFe^{+2}}$= 1.75 mmol MnO₂

Then we convert MnO₂ to Mn₃O₄, using the reaction:

• 3MnO₂ → Mn₃O₄ + O₂

• 1.75 mmol MnO₂ * $\frac{1mmolMn_3O_4}{3mmolMnO_2}$ = 0.583 mmol Mn₃O₄

Finally we convert Mn₃O₄ moles to grams:

• 0.583 mmol Mn₃O₄ * 228.82 mg/mmol = 133.40 mg Mn₃O₄

And calculate the percent

• 0.2 g = 200 mg

• 133.40 / 200 * 100% = 66.7%