`Al_(2)O_(3)` is reduced by electrolysis at low potentials and high current. If `4.0 xx 10^(4)` amperes of currernt is passed through molten `Al_(2)O_(3)` for 6 hours, what mass of aluminium is produced? (Assume 100 % current efficiency, At. Mass of `Al = 27 u`)

Options

A. `8.1 xx 10^(4) g`
B. `2.4 xx 10^(5) g`
C. `1.3 xx 10^(4) g`
D. `9.0 xx 10^(3) g`

Explanation

To solve the problem of how much aluminum is produced by the electrolysis of molten Al₂O₃ when a current of 4.0 x 10⁴ amperes is passed for 6 hours, we can follow these steps:

### Step 1: Write the Half-Reaction
The reduction of aluminum ions from Al₂O₃ can be represented by the half-reaction:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al (s)} \]

### Step 2: Determine the Equivalent Weight
The equivalent weight (E) of aluminum can be calculated using the formula:
\[ E = \frac{\text{Molar Mass}}{\text{Valency}} \]
Given that the atomic mass of aluminum (Al) is 27 g/mol and its valency is 3 (since one Al atom gains 3 electrons), we have:
\[ E = \frac{27 \text{ g/mol}}{3} = 9 \text{ g/equiv} \]

### Step 3: Calculate Total Charge (Q)
Using the formula for charge:
\[ Q = I \times t \]
where:
- \( I = 4.0 \times 10^4 \) A (current)
- \( t = 6 \text{ hours} = 6 \times 3600 \text{ seconds} = 21600 \text{ seconds} \)

Calculating \( Q \):
\[ Q = 4.0 \times 10^4 \text{ A} \times 21600 \text{ s} = 864000000 \text{ C} \]

### Step 4: Calculate the Number of Moles of Electrons
Using Faraday's constant (F = 96500 C/mol), we can find the number of moles of electrons (n):
\[ n = \frac{Q}{F} = \frac{864000000 \text{ C}}{96500 \text{ C/mol}} \approx 8965.4 \text{ mol} \]

### Step 5: Calculate the Moles of Aluminum Produced
From the half-reaction, we see that 3 moles of electrons produce 1 mole of aluminum. Therefore, the moles of aluminum produced (n(Al)) is:
\[ n(\text{Al}) = \frac{n}{3} = \frac{8965.4}{3} \approx 2988.5 \text{ mol} \]

### Step 6: Calculate the Mass of Aluminum Produced
Finally, we can calculate the mass of aluminum produced using the molar mass:
\[ \text{Mass} = n(\text{Al}) \times \text{Molar Mass} = 2988.5 \text{ mol} \times 27 \text{ g/mol} \approx 80685 \text{ g} \approx 81 \text{ kg} \]

### Final Answer
The mass of aluminum produced is approximately **81 kg**.

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