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 first need to write and balance the chemical equation for the reaction.

### Step 1: Write the balanced chemical equation
The reaction between iron (Fe) and oxygen (O2) to form iron(III) oxide (Fe2O3) can be represented as:
\[ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \]

### Step 2: Identify the stoichiometric ratios
From the balanced equation, we can see that:
- 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3.

### Step 3: Determine the limiting reagent
We have:
- 0.5 moles of Fe
- 0.5 moles of O2

To find the limiting reagent, we need to determine how much O2 is required for the available Fe:
- For 4 moles of Fe, 3 moles of O2 are needed.
- Therefore, for 0.5 moles of Fe, the required O2 is calculated as:
\[ \text{Required O2} = \left(\frac{3 \text{ moles O2}}{4 \text{ moles Fe}}\right) \times 0.5 \text{ moles Fe} = 0.375 \text{ moles O2} \]

Since we have 0.5 moles of O2 available, O2 is in excess and Fe is the limiting reagent.

### Step 4: Calculate the amount of Fe2O3 produced
Now, we can use the limiting reagent (Fe) to find out how much Fe2O3 will be produced:
- From the balanced equation, 4 moles of Fe produce 2 moles of Fe2O3.
- Therefore, for 0.5 moles of Fe:
\[ \text{Fe2O3 produced} = \left(\frac{2 \text{ moles Fe2O3}}{4 \text{ moles Fe}}\right) \times 0.5 \text{ moles Fe} = 0.25 \text{ moles Fe2O3} \]

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

---