How much electricity in terms of Faradays is required to produce ?
`81 ` g of Al from molten `Al_(2) O_(3)`

Explanation

To determine how much electricity in terms of Faradays is required to produce 81 g of aluminum from molten Al₂O₃, we can follow these steps:

### Step 1: Determine the molar mass of aluminum (Al)
The molar mass of aluminum (Al) is approximately 27 g/mol.

### Step 2: Calculate the number of moles of aluminum produced
To find the number of moles of aluminum produced from 81 g, we can use the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

Substituting the values:

\[ \text{Number of moles of Al} = \frac{81 \, \text{g}}{27 \, \text{g/mol}} = 3 \, \text{mol} \]

### Step 3: Determine the number of electrons required
From the electrochemical reaction for the reduction of Al₂O₃ to aluminum, we can see that:

\[ \text{Al}_2\text{O}_3 + 6e^- \rightarrow 2\text{Al} + 3\text{O}^{2-} \]

This indicates that to produce 2 moles of Al, 6 moles of electrons are required. Therefore, to produce 3 moles of Al, the number of moles of electrons required can be calculated as follows:

\[ \text{Electrons required} = \left(\frac{6 \, \text{mol e}^-}{2 \, \text{mol Al}}\right) \times 3 \, \text{mol Al} = 9 \, \text{mol e}^- \]

### Step 4: Convert moles of electrons to Faradays
Since 1 mole of electrons corresponds to 1 Faraday, the total number of Faradays required is equal to the number of moles of electrons:

\[ \text{Faradays required} = 9 \, \text{mol e}^- = 9 \, \text{Faradays} \]

### Conclusion
Thus, the amount of electricity required to produce 81 g of aluminum from molten Al₂O₃ is **9 Faradays**.

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