The desired amount of charge for obtaining one mole of Al from `Al^(3+)`

Options

A. `3xx96500C`
B. `96500C`
C. `(96500)/(3)C`
D. `(96500)/(2)C`

Explanation

To determine the desired amount of charge for obtaining one mole of aluminum (Al) from aluminum ions (\(Al^{3+}\)), we can follow these steps:

### Step 1: Understand the Reduction Process
Aluminum ions (\(Al^{3+}\)) need to gain electrons to be reduced to aluminum metal (Al). The half-reaction for this process is:
\[ Al^{3+} + 3e^- \rightarrow Al \]
This indicates that one mole of \(Al^{3+}\) requires three moles of electrons to be reduced to one mole of aluminum.

### Step 2: Calculate the Moles of Electrons Required
From the half-reaction, we see that to produce one mole of aluminum, we need three moles of electrons:
\[ \text{Moles of electrons required} = 3 \text{ moles} \]

### Step 3: Use Faraday's Constant
The charge required to transfer one mole of electrons is given by Faraday's constant, which is approximately \(96500 \, C/mol\). Therefore, the total charge (Q) required for three moles of electrons can be calculated as:
\[ Q = \text{Moles of electrons} \times \text{Faraday's constant} \]
Substituting the values:
\[ Q = 3 \, \text{moles} \times 96500 \, C/mol \]

### Step 4: Perform the Calculation
Now, calculate the total charge:
\[ Q = 3 \times 96500 = 289500 \, C \]

### Conclusion
The desired amount of charge for obtaining one mole of aluminum from \(Al^{3+}\) is:
\[ Q = 289500 \, C \]