The quantity of charge required to obtain one mole of aluminium from `Al_(2)0_(3)` is

Options

A. 1 F
B. 6 F
C. 3 F
D. 2 F

Explanation

To find the quantity of charge required to obtain one mole of aluminum from Al₂O₃, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The decomposition of aluminum oxide (Al₂O₃) to produce aluminum (Al) can be represented by the following half-reaction:
\[ Al_2O_3 \rightarrow 2Al + \frac{3}{2}O_2 \]

2. **Determine the Moles of Aluminum Produced**:
From the equation, we see that 1 mole of Al₂O₃ produces 2 moles of Al.

3. **Write the Half-Reaction for Aluminum Reduction**:
The reduction half-reaction for aluminum can be written as:
\[ Al^{3+} + 3e^- \rightarrow Al \]
This indicates that to produce 1 mole of Al, 3 moles of electrons (3e⁻) are required.

4. **Calculate the Total Charge Required**:
The total charge (Q) required can be calculated using Faraday's constant (F), which is approximately 96500 C/mol of electrons. For 1 mole of Al, we need 3 moles of electrons:
\[ Q = n \times F = 3 \times 96500 \, \text{C} = 289500 \, \text{C} \]

5. **Final Answer**:
Therefore, the quantity of charge required to obtain one mole of aluminum from Al₂O₃ is:
\[ 3F \]

### Summary:
The quantity of charge required to obtain one mole of aluminum from Al₂O₃ is **3F**.