To solve the problem of calculating the mass of aluminum produced from the electrolysis of molten `Al2O3`, we will follow these steps:
### Step 1: Write the half-reaction for aluminum reduction
The reduction of aluminum ions can be represented by the half-reaction:
\[
\text{Al}^{3+} + 3e^- \rightarrow \text{Al}
\]
This indicates that 1 mole of aluminum is produced by the gain of 3 moles of electrons.
### Step 2: Calculate the total charge (Q) passed through the system
The total charge can be calculated using the formula:
\[
Q = I \times T
\]
where:
- \( I \) is the current in amperes (A),
- \( T \) is the time in seconds (s).
Given:
- \( I = 4.0 \times 10^4 \, \text{A} \)
- \( T = 6 \, \text{hours} = 6 \times 60 \times 60 \, \text{s} = 21600 \, \text{s} \)
Calculating \( Q \):
\[
Q = 4.0 \times 10^4 \, \text{A} \times 21600 \, \text{s} = 864000000 \, \text{C}
\]
### Step 3: Calculate the number of moles of electrons (n)
Using Faraday's constant, which is approximately \( 96500 \, \text{C/mol} \), we can find the number of moles of electrons:
\[
n = \frac{Q}{F}
\]
where \( F \) is Faraday's constant.
Calculating \( n \):
\[
n = \frac{864000000 \, \text{C}}{96500 \, \text{C/mol}} \approx 8956.5 \, \text{mol}
\]
### Step 4: Calculate the moles of aluminum produced
From the half-reaction, we know that 3 moles of electrons produce 1 mole of aluminum. Therefore, the moles of aluminum produced can be calculated as:
\[
\text{Moles of Al} = \frac{n}{3}
\]
Calculating the moles of aluminum:
\[
\text{Moles of Al} = \frac{8956.5 \, \text{mol}}{3} \approx 2985.5 \, \text{mol}
\]
### Step 5: Calculate the mass of aluminum produced
The mass of aluminum can be calculated using the formula:
\[
\text{Mass} = \text{Moles} \times \text{Molar mass}
\]
Given the molar mass of aluminum is \( 27 \, \text{g/mol} \):
\[
\text{Mass} = 2985.5 \, \text{mol} \times 27 \, \text{g/mol} \approx 80400.5 \, \text{g}
\]
### Final Answer
The mass of aluminum produced is approximately:
\[
\text{Mass} \approx 80400.5 \, \text{g}
\]
