The number of moles of `Fe_(2)O_(3)` formed when 0.5 moles of `O_(2)` and `0.5` moles of Fe are allowed to react are

Options

A. `0.25`
B. `0.5`
C. `1//3`
D. `0.125`

Explanation

To determine the number of moles of `Fe_(2)O_(3)` formed when 0.5 moles of `O_(2)` and 0.5 moles of `Fe` are allowed to react, we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The formation of iron(III) oxide (`Fe_(2)O_(3)`) from iron (`Fe`) and oxygen (`O_(2)`) can be represented by the following unbalanced equation:
\[ \text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 \]

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:
\[ 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \]

### Step 2: Identify the Stoichiometric Ratios
From the balanced equation, we can see the stoichiometric ratios:
- 4 moles of `Fe` react with 3 moles of `O_(2)` to produce 2 moles of `Fe_(2)O_(3)`.

### Step 3: Determine the Limiting Reagent
We have:
- 0.5 moles of `Fe`
- 0.5 moles of `O_(2)`

Now, we will calculate how much of each reactant is needed based on the stoichiometric ratios.

For `Fe`:
- The ratio from the balanced equation is \( \frac{4 \text{ moles Fe}}{2 \text{ moles Fe}_2\text{O}_3} \).
- Therefore, for 0.5 moles of `Fe`, the moles of `Fe_(2)O_(3)` produced would be:
\[ \frac{0.5 \text{ moles Fe}}{4 \text{ moles Fe}} \times 2 \text{ moles Fe}_2\text{O}_3 = 0.25 \text{ moles Fe}_2\text{O}_3 \]

For `O_(2)`:
- The ratio from the balanced equation is \( \frac{3 \text{ moles O}_2}{2 \text{ moles Fe}_2\text{O}_3} \).
- Therefore, for 0.5 moles of `O_(2)`, the moles of `Fe_(2)O_(3)` produced would be:
\[ \frac{0.5 \text{ moles O}_2}{3 \text{ moles O}_2} \times 2 \text{ moles Fe}_2\text{O}_3 \approx 0.33 \text{ moles Fe}_2\text{O}_3 \]

### Step 4: Identify the Limiting Reagent
From the calculations:
- `Fe` can produce 0.25 moles of `Fe_(2)O_(3)`.
- `O_(2)` can produce approximately 0.33 moles of `Fe_(2)O_(3)`.

Since `Fe` produces fewer moles of `Fe_(2)O_(3)`, it is the limiting reagent.

### Step 5: Calculate the Moles of `Fe_(2)O_(3)` Formed
The number of moles of `Fe_(2)O_(3)` formed is determined by the limiting reagent:
- Therefore, the number of moles of `Fe_(2)O_(3)` formed is 0.25 moles.

### Final Answer
The number of moles of `Fe_(2)O_(3)` formed is **0.25 moles**.

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