A 3.00 g sample containing `Fe_(3)O_(4), Fe_(2)O_(3)` and an inert impure substance, is treated with excess of KI solution in presence of dilute `H_(2)SO_(4)` . The entire iron is converted into `Fe^(2+)` along with the liberation of iodine. The resulting solution is diluted to 100 mL. A 20 mL of the dilute solution requires 11.0 mL of 0.5 M `Na_(2)S_(2)O_(3)` solution to reduce the iodine, present. A 50 mL of the dilute solution after complete extraction of the iodine required 12.80 mL of 0.25 M `KMnO_(4)` solution in dilute `H_(2)SO_(4)` medium for the oxidation of `Fe^(2+)` . Calculate the % of `Fe_(2)O_(3) "and" Fe_(3)O_(4)` in the original sample.

Explanation

To solve the problem step by step, we need to determine the percentage of `Fe_(2)O_(3)` and `Fe_(3)O_(4)` in the original sample based on the given data.

### Step 1: Understanding the Reaction
The sample contains `Fe_(3)O_(4)` and `Fe_(2)O_(3)`, which react with potassium iodide (KI) in the presence of dilute sulfuric acid (`H_(2)SO_(4)`) to produce `Fe^(2+)` ions and liberate iodine (`I_2`).

### Step 2: Calculate Moles of Iodine Liberated
From the titration data:
- A 20 mL sample of the diluted solution requires 11.0 mL of 0.5 M `Na_(2)S_(2)O_(3)` to reduce the iodine.
...