In the reaction
`8Al+3Fe_(3)O_(4)rarr 4Al_(2)O_(3)+9Fe`
the number of electrons transferred from the reductant to the oxidant is

Options

A. `8`
B. `4`
C. `16`
D. `24`

Explanation

To determine the number of electrons transferred from the reductant (Aluminum, Al) to the oxidant (Iron in Fe3O4) in the given reaction:

**Reaction:**
\[ 8 \text{Al} + 3 \text{Fe}_3\text{O}_4 \rightarrow 4 \text{Al}_2\text{O}_3 + 9 \text{Fe} \]

**Step 1: Identify the oxidation states of the elements involved.**
- In elemental form, the oxidation state of Aluminum (Al) is 0.
- In Aluminum oxide (\( \text{Al}_2\text{O}_3 \)), the oxidation state of Al is +3.
- In \( \text{Fe}_3\text{O}_4 \), the oxidation state of Iron (Fe) can be determined as follows:
- The overall charge of \( \text{Fe}_3\text{O}_4 \) is neutral (0).
- Let the oxidation state of Fe be \( x \). Then, we have:
\[ 3x + 4(-2) = 0 \]
\[ 3x - 8 = 0 \]
\[ 3x = 8 \]
\[ x = \frac{8}{3} \approx +2.67 \]
- Therefore, the average oxidation state of Fe in \( \text{Fe}_3\text{O}_4 \) is approximately +2.67.

**Step 2: Determine the change in oxidation states.**
- For Aluminum:
- Change from 0 (in Al) to +3 (in \( \text{Al}_2\text{O}_3 \)).
- Each Al atom loses 3 electrons.

- For Iron:
- Change from +2.67 (in \( \text{Fe}_3\text{O}_4 \)) to 0 (in elemental Fe).
- Each Fe atom gains approximately 2.67 electrons.

**Step 3: Calculate the total number of electrons transferred.**
- Since there are 8 Aluminum atoms:
\[ \text{Total electrons lost by Al} = 8 \times 3 = 24 \text{ electrons} \]

**Step 4: Conclusion**
- The total number of electrons transferred from the reductant (Al) to the oxidant (Fe) in the reaction is **24 electrons**.

**Final Answer:**
The number of electrons transferred from the reductant to the oxidant is **24**.

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