Question

A 0.500 g sample of an impure iron ore is converted entirely into Fe2+, which is then titrated with KMnO4 (source of MnO4- ions) in the following reaction: 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O It requires 32.0 mL of 0.0200 M KMnO4 to react completely with all the Fe2+ in the solution. What is the percent by mass of iron in the ore?

Answer 1

Explanation:

The given balanced reaction equation is as follows.

           $5Fe^{2+} + MnO^{-}_{4} + 8H^{+} \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_{2}O$

The given data is as follows.

      Mass of solution (that is sample) = 0.5 g

Therefore, calculate the mass of solute as follows.

      $0.032 L \times \frac{0.02 mol MnO^{-}_{4}}{1 L} \times \frac{5 mol Fe^{2+}}{1 mol MnO^{-}_{4}} \times \frac{1 mol Fe}{1 mol Fe^{3+}} \times \frac{55.845 g Fe}{1 mol Fe}$

            = 0.178704 g Fe

Now, calculate the percent by mass Fe in the ore as follows.

      Percent by mass of Fe in the ore = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100%$

                              = $\frac{0.178704 g}{0.5 g} \times 100%$

                              = 35.7%

Thus, we can conclude that percent by mass of iron in the ore is 35.7%.