When `Fe_(0.93)O` is heated in presence of oxygen, it converts to `Fe_2O_3`. The number of correct statement/s from the following is _____________

Options

A. A. The equivalent weight of `Fe_(0.93)O` is Molecular weight//0.79
B. The number of moles of `Fe^(2+ )` and `Fe^(3+)` in 1 mole of `Fe_(0.93)O` is 0.79 and 0.14 respectively
C. `Fe_(0.93)O` is metal deficient with lattice comprising of cubic closed packed arrangement of `O^(2–)` ions
D. The % of `Fe^(+2)` and `Fe^(3+)` in `Fe_(0.93)O` is 85% and 15% respectively

Answer

To solve the question regarding the conversion of `Fe_(0.93)O` to `Fe_2O_3` when heated in the presence of oxygen, we will analyze the statements provided and determine their correctness step by step.

### Step-by-Step Solution:

1. **Understanding the Composition of `Fe_(0.93)O`:**
- The formula `Fe_(0.93)O` indicates that there are 0.93 moles of iron (Fe) for every mole of oxygen (O). This suggests that the compound is iron-deficient, meaning it has fewer iron atoms than expected in a stoichiometric oxide.

2. **Oxidation States:**
- In `Fe_(0.93)O`, the oxidation state of oxygen is -2. Let the oxidation state of iron be x. The overall charge balance gives us:
\[ 0.93x + (-2) = 0 \implies 0.93x = 2 \implies x = \frac{2}{0.93} \approx +2.15 \]
- This indicates that iron is in a mixed oxidation state, primarily +2.

3. **Conversion to `Fe_2O_3`:**
- When `Fe_(0.93)O` is heated in the presence of oxygen, it converts to `Fe_2O_3`. In `Fe_2O_3`, the oxidation state of iron is +3.

4. **Calculating Moles of Iron in `Fe_2O_3`:**
- The conversion can be represented as:
\[ 4Fe_{0.93}O + O_2 \rightarrow 2Fe_2O_3 \]
- From `Fe_(0.93)O`, we have 0.93 moles of Fe. The total moles of Fe in `Fe_2O_3` will be:
\[ 2 \times 2 = 4 \text{ moles of Fe} \]
- To find the moles of `Fe_(0.93)O` needed to produce 2 moles of `Fe_2O_3`, we can set up the equation:
\[ 0.93n = 4 \implies n = \frac{4}{0.93} \approx 4.30 \text{ moles of } Fe_{0.93}O \]

5. **Determining the Correct Statements:**
- We need to evaluate the statements provided in the question regarding the properties of `Fe_(0.93)O` and `Fe_2O_3`.
- **Statement 1:** The equivalent weight of `Fe_(0.93)O` can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
where n is the number of moles of electrons transferred. In this case, n is approximately 0.79.
- **Statement 2:** The percentage of iron in `Fe_(0.93)O` can be calculated as:
\[ \text{Percentage of Fe} = \left( \frac{0.93 \times 55.85}{(0.93 \times 55.85) + (1 \times 16)} \right) \times 100 \]
This should give approximately 85% for Fe and 15% for O.

6. **Final Evaluation:**
- After evaluating the statements based on the calculations, we find that both statements are correct.

### Conclusion:
The number of correct statements is **2**.