0.7 g sample of iron ore was dissolved in acid. Iron was reduced to `+2` state and it required 50 " mL of " `(M)/(50)` `KMnO_(4)` solution for titration. The percentage of Fe and `Fe_(3)O_(4)` in the ore is

Options

A. `40%Fe,55.24%,Fe_(3)O_(4)`
B. `55.24%Fe,40%Fe_(3)O_(4)`
C. `8%Fe,11%Fe_(3)O_(4)`
D. `11%Fe,8%Fe_(3)O_(4)`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the moles of KMnO4 used in the titration
The molarity of the KMnO4 solution is given as \( \frac{1}{50} \) M, and the volume used is 50 mL.

\[ \text{Moles of KMnO}_4 = \text{Molarity} \times \text{Volume (in L)} = \frac{1}{50} \times \frac{50}{1000} = 0.001 \text{ moles} \]

### Step 2: Calculate the milliequivalents of KMnO4
The n-factor for KMnO4 in this reaction is 5 (since it reduces from +7 to +2, gaining 5 electrons).

\[ \text{Milliequivalents of KMnO}_4 = \text{Moles} \times \text{n-factor} \times 1000 = 0.001 \times 5 \times 1000 = 5 \text{ mEq} \]

### Step 3: Relate the milliequivalents of KMnO4 to Fe²⁺
From the balanced reaction, we see that 1 equivalent of KMnO4 reacts with 5 equivalents of Fe²⁺. Therefore, the milliequivalents of Fe²⁺ will also be 5 mEq.

### Step 4: Calculate the mass of Fe in the sample
The molar mass of Fe is 56 g/mol. To find the mass of Fe corresponding to 5 mEq:

\[ \text{Mass of Fe} = \text{Milliequivalents} \times \frac{\text{Molar mass}}{1000} = 5 \times \frac{56}{1000} = 0.280 \text{ g} \]

### Step 5: Calculate the percentage of Fe in the iron ore sample
The mass of the iron ore sample is 0.7 g. The percentage of Fe is calculated as follows:

\[ \text{Percentage of Fe} = \left( \frac{\text{Mass of Fe}}{\text{Mass of sample}} \right) \times 100 = \left( \frac{0.280}{0.7} \right) \times 100 = 40\% \]

### Step 6: Calculate the percentage of Fe₃O₄ in the sample
In Fe₃O₄, there are 3 moles of Fe. The molar mass of Fe₃O₄ is calculated as follows:

\[ \text{Molar mass of Fe}_3\text{O}_4 = 3 \times 56 + 4 \times 16 = 168 + 64 = 232 \text{ g/mol} \]

Now, we can find the mass of Fe₃O₄ that corresponds to the 40% of Fe:

\[ \text{Mass of Fe from Fe}_3\text{O}_4 = \text{Percentage of Fe} \times \text{Mass of Fe}_3\text{O}_4 = \frac{40}{100} \times 232 = 92.8 \text{ g} \]

To find the percentage of Fe₃O₄ in the sample:

\[ \text{Percentage of Fe}_3\text{O}_4 = \left( \frac{\text{Mass of Fe from Fe}_3\text{O}_4}{\text{Mass of sample}} \right) \times 100 = \left( \frac{92.8}{168} \right) \times 100 \approx 55.24\% \]

### Final Result
The percentage of Fe in the ore is **40%** and the percentage of Fe₃O₄ in the ore is approximately **55.24%**.

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