The composition of a sample of wustite is `Fe_(0.93)O_(1.0)`. What is the percentage of iron present as `Fe^(3+)` ?

Options

A. `15%`
B. `25%`
C. `35%`
D. `45%`

Explanation

To find the percentage of iron present as \( \text{Fe}^{3+} \) in the compound \( \text{Fe}_{0.93}\text{O}_{1.0} \), we can follow these steps:

### Step 1: Understand the composition
The formula \( \text{Fe}_{0.93}\text{O}_{1.0} \) indicates that there are 0.93 moles of iron (Fe) and 1.0 mole of oxygen (O) in the compound.

### Step 2: Determine the total charge
Since oxygen typically has a charge of \( -2 \) (for \( \text{O}^{2-} \)), the total negative charge from oxygen in this compound is:
\[ \text{Total charge from O} = 1.0 \times (-2) = -2 \]

### Step 3: Set up the charge balance equation
Let \( x \) be the amount of \( \text{Fe}^{3+} \) present in the compound. Then, the amount of \( \text{Fe}^{2+} \) will be \( 0.93 - x \). The charges can be expressed as:
- Charge from \( \text{Fe}^{3+} \): \( 3x \)
- Charge from \( \text{Fe}^{2+} \): \( 2(0.93 - x) = 1.86 - 2x \)

The total charge must equal the charge from oxygen:
\[ 3x + (1.86 - 2x) = 2 \]

### Step 4: Solve the equation
Now, simplify and solve for \( x \):
\[ 3x + 1.86 - 2x = 2 \]
\[ x + 1.86 = 2 \]
\[ x = 2 - 1.86 = 0.14 \]

Thus, the amount of \( \text{Fe}^{3+} \) is \( 0.14 \).

### Step 5: Calculate the percentage of \( \text{Fe}^{3+} \)
To find the percentage of iron that is in the \( \text{Fe}^{3+} \) state, use the formula:
\[ \text{Percentage of } \text{Fe}^{3+} = \left( \frac{\text{Amount of } \text{Fe}^{3+}}{\text{Total amount of Fe}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage of } \text{Fe}^{3+} = \left( \frac{0.14}{0.93} \right) \times 100 \]

### Step 6: Perform the calculation
Calculating the above expression:
\[ \text{Percentage of } \text{Fe}^{3+} \approx 15.05\% \]

Rounding this to two decimal places gives approximately \( 15\% \).

### Final Answer
The percentage of iron present as \( \text{Fe}^{3+} \) in the compound \( \text{Fe}_{0.93}\text{O}_{1.0} \) is **15%**.

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