When non stoiciometric compound `Fe_(0.95)O` is heated in presence of oxygen then it convents into `Fe_(2)O_(3)`. Which of the following statement is correct?

Options

A. Equivalent weight of `Fe_(0.95)O` is `(MB)/0.5`. Where `M_(B)` is molecular weight of `Fe_(0.95)O`
B. The number of moles of `Fe^(+3)` and `Fe^(+2)M` 1 mole `Fe_(0.95)O` are 0.1 are 0.85 respectively
C. The number of moles of `Fe^(+3)`, and `Fe^(+2)` in 1 mole of `Fe_(0.95)O` are 0.85 and 0.1 respectively.
D. The % composition of `Fe^(+2)` and `Fe^(+3)` in the non stoichiometric compound is 89.47% and 10.53 %

Answer

To solve the question regarding the conversion of the non-stoichiometric compound `Fe_(0.95)O` into `Fe_(2)O_(3)` when heated in the presence of oxygen, we will analyze the composition of the compound and the resulting changes. Here’s the step-by-step solution:

### Step 1: Understand the Composition of `Fe_(0.95)O`
The compound `Fe_(0.95)O` indicates that there are 0.95 moles of iron (Fe) for every mole of oxygen (O). This means that the iron in this compound is in two oxidation states: Fe²⁺ (ferrous) and Fe³⁺ (ferric).

### Step 2: Set Up the Charge Balance
Let:
- The number of moles of Fe²⁺ be \( y \).
- The number of moles of Fe³⁺ be \( 0.95 - y \).

Since the compound is neutral, the total positive charge must equal the total negative charge from the oxygen ions. The charge balance can be set up as follows:

\[ 2y + 3(0.95 - y) = 2 \]

### Step 3: Solve for \( y \)
Expanding the equation gives:

\[ 2y + 2.85 - 3y = 2 \]

Rearranging this, we get:

\[ - y + 2.85 = 2 \]

Thus,

\[ y = 2.85 - 2 = 0.85 \]

### Step 4: Calculate Moles of Fe³⁺
Now, substituting \( y \) back to find the moles of Fe³⁺:

\[ Fe^{3+} = 0.95 - 0.85 = 0.10 \]

### Step 5: Calculate Percentages of Fe²⁺ and Fe³⁺
Now we can calculate the percentage of each ion in the compound:

1. **Percentage of Fe²⁺:**
\[ \text{Percentage of Fe}^{2+} = \left( \frac{0.85 \times 56}{0.95 \times 56} \right) \times 100 = \left( \frac{0.85}{0.95} \right) \times 100 \approx 89.47\% \]

2. **Percentage of Fe³⁺:**
\[ \text{Percentage of Fe}^{3+} = \left( \frac{0.10 \times 56}{0.95 \times 56} \right) \times 100 = \left( \frac{0.10}{0.95} \right) \times 100 \approx 10.53\% \]

### Step 6: Determine Correct Statements
Now we can analyze the statements provided in the question. Based on our calculations:
- The percentage of Fe²⁺ is approximately 89.47%.
- The percentage of Fe³⁺ is approximately 10.53%.
- The mole ratio of Fe³⁺ to Fe²⁺ is 0.1 to 0.85.

### Conclusion
From the calculations, we find that:
- The equivalent mass of `Fe_(0.95)O` is given by the formula \( \text{Molecular mass} / 0.5 \), which is correct.
- The moles of Fe²⁺ and Fe³⁺ are correctly calculated.

Thus, the correct statements are those that match our findings regarding the percentages and the equivalent mass.