The mass percentage of `Fe^(3+)` ion present in `Fe_(0.93), O_(1.00)` is

Options

A. `15%`
B. `5.5%`
C. `10.0%`
D. `11.5%`

Answer

To find the mass percentage of `Fe^(3+)` ion present in `Fe_(0.93)O_(1.00)`, we can follow these steps:

### Step 1: Determine the oxidation states of iron
In the compound `Fe_(0.93)O_(1.00)`, iron can exist in two oxidation states: `Fe^(2+)` and `Fe^(3+)`. We will denote the amount of `Fe^(3+)` as `X` and the amount of `Fe^(2+)` as `0.93 - X`.

### Step 2: Set up the charge balance equation
To maintain electrical neutrality, the total positive charge from the iron ions must equal the negative charge from the oxygen ions. The charge from the oxygen is `-2` (since there is one `O` with a charge of `-2`), so we can set up the equation:
\[ (0.93 - X) \cdot 2 + X \cdot 3 = 2 \]
This simplifies to:
\[ 1.86 - 2X + 3X = 2 \]
\[ 1.86 + X = 2 \]

### Step 3: Solve for X
Now, we can solve for `X`:
\[ X = 2 - 1.86 = 0.14 \]
This means there are `0.14` moles of `Fe^(3+)`.

### Step 4: Calculate the moles of `Fe^(2+)`
The moles of `Fe^(2+)` can be calculated as:
\[ 0.93 - X = 0.93 - 0.14 = 0.79 \]

### Step 5: Calculate the molar mass of the compound
Next, we calculate the molar mass of `Fe_(0.93)O_(1.00)`:
- Molar mass of `Fe` = 56 g/mol
- Molar mass of `O` = 16 g/mol

Thus, the molar mass of the compound is:
\[ (0.93 \times 56) + (1.00 \times 16) = 52.08 + 16 = 68.08 \text{ g/mol} \]

### Step 6: Calculate the mass of `Fe^(3+)`
Now, we calculate the mass of `Fe^(3+)`:
\[ \text{Mass of } Fe^{3+} = 0.14 \text{ moles} \times 56 \text{ g/mol} = 7.84 \text{ g} \]

### Step 7: Calculate the mass percentage of `Fe^(3+)`
Finally, we can find the mass percentage of `Fe^(3+)` in the compound:
\[ \text{Mass percentage of } Fe^{3+} = \left( \frac{7.84 \text{ g}}{68.08 \text{ g}} \right) \times 100 \approx 11.5\% \]

### Conclusion
The mass percentage of `Fe^(3+)` ion present in `Fe_(0.93)O_(1.00)` is approximately **11.5%**.

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