How many faradays/coulombs are required to produce (i) 20.0 g of calcium from molten `CaCl_(2)`?
(ii) 40.0 g of aluminium from molten `Al_(2)O_(3)`?

Explanation

To solve the question, we need to calculate the amount of Faradays (or Coulombs) required to produce specified amounts of calcium and aluminum from their respective molten salts.

### Part (i): Producing 20.0 g of Calcium from Molten CaCl₂

1. **Determine the molar mass of Calcium (Ca)**:
- The atomic weight of calcium (Ca) is 40 g/mol.

2. **Calculate the number of moles of Calcium**:
\[ \text{Moles of Ca} = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{40 \, \text{g/mol}} = 0.5 \, \text{mol} \]

3. **Determine the number of electrons required for the reduction of Calcium**:
- The reaction for the reduction of calcium from Ca²⁺ is:
\[ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \]
- This means that 1 mole of Ca requires 2 moles of electrons.

4. **Calculate the total moles of electrons required**:
\[ \text{Moles of electrons} = 0.5 \, \text{mol Ca} \times 2 \, \text{mol e}^- / \text{mol Ca} = 1.0 \, \text{mol e}^- \]

5. **Convert moles of electrons to Faradays**:
- Since 1 mole of electrons corresponds to 1 Faraday, we need 1 Faraday to produce 20.0 g of calcium.

6. **Convert Faradays to Coulombs**:
\[ \text{Coulombs} = \text{Faradays} \times 96500 \, \text{C/F} = 1 \, \text{F} \times 96500 \, \text{C/F} = 96500 \, \text{C} \]

### Part (ii): Producing 40.0 g of Aluminum from Molten Al₂O₃

1. **Determine the molar mass of Aluminum (Al)**:
- The atomic weight of aluminum (Al) is 27 g/mol.

2. **Calculate the number of moles of Aluminum**:
\[ \text{Moles of Al} = \frac{40.0 \, \text{g}}{27 \, \text{g/mol}} \approx 1.481 \, \text{mol} \]

3. **Determine the number of electrons required for the reduction of Aluminum**:
- The reaction for the reduction of aluminum from Al³⁺ is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
- This means that 1 mole of Al requires 3 moles of electrons.

4. **Calculate the total moles of electrons required**:
\[ \text{Moles of electrons} = 1.481 \, \text{mol Al} \times 3 \, \text{mol e}^- / \text{mol Al} \approx 4.443 \, \text{mol e}^- \]

5. **Convert moles of electrons to Faradays**:
- Therefore, we need approximately 4.443 Faradays to produce 40.0 g of aluminum.

6. **Convert Faradays to Coulombs**:
\[ \text{Coulombs} = 4.443 \, \text{F} \times 96500 \, \text{C/F} \approx 428460 \, \text{C} \]

### Summary of Results:
- (i) To produce 20.0 g of calcium, **1 Faraday** or **96500 Coulombs** is required.
- (ii) To produce 40.0 g of aluminum, approximately **4.443 Faradays** or **428460 Coulombs** is required.