The mass percentage of iron present as Fe(III) in `Fe_(-.93) O_(1.0)` is

Options

A. `8.3%`
B. `9.6%`
C. `11.5%`
D. `17.7%`

Explanation

To find the mass percentage of iron present as Fe(III) in the compound Fe₋₉₃O₁, we will follow these steps:

### Step 1: Determine the Molar Mass of the Compound
The molar mass of Fe is approximately 56 g/mol, and the molar mass of O is approximately 16 g/mol. The formula given is Fe₋₉₃O₁, which means we have 0.93 moles of Fe and 1 mole of O.

- Molar mass of Fe₋₉₃ = 0.93 × 56 g/mol = 52.08 g
- Molar mass of O = 1 × 16 g/mol = 16 g
- Total molar mass of Fe₋₉₃O₁ = 52.08 g + 16 g = 68.08 g

### Step 2: Set Up the Charge Balance Equation
Let \( x \) be the number of Fe(III) ions (Fe³⁺). The remaining iron will be in the form of Fe(II) ions (Fe²⁺), which will be \( 0.93 - x \).

The charge balance for the compound can be set up as follows:
- The total positive charge from Fe³⁺ is \( 3x \).
- The total positive charge from Fe²⁺ is \( 2(0.93 - x) \).
- The total negative charge from O is \( -2 \) (since there is 1 O²⁻).

Setting up the equation:
\[ 3x + 2(0.93 - x) = 2 \]

### Step 3: Solve for \( x \)
Expanding the equation:
\[ 3x + 1.86 - 2x = 2 \]
\[ x + 1.86 = 2 \]
\[ x = 2 - 1.86 \]
\[ x = 0.14 \]

So, there are 0.14 moles of Fe³⁺.

### Step 4: Calculate the Mass of Fe³⁺
To find the mass of Fe³⁺:
\[ \text{Mass of Fe³⁺} = 0.14 \, \text{moles} \times 56 \, \text{g/mol} = 7.84 \, \text{g} \]

### Step 5: Calculate the Mass Percentage of Fe³⁺
Now, we can calculate the mass percentage of Fe³⁺ in the compound:
\[ \text{Mass percentage of Fe³⁺} = \left( \frac{\text{Mass of Fe³⁺}}{\text{Total mass of compound}} \right) \times 100 \]
\[ \text{Mass percentage of Fe³⁺} = \left( \frac{7.84 \, \text{g}}{68.08 \, \text{g}} \right) \times 100 \]
\[ \text{Mass percentage of Fe³⁺} \approx 11.5\% \]

### Final Answer
The mass percentage of iron present as Fe(III) in Fe₋₉₃O₁ is approximately **11.5%**.

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