To find the mass of aluminum produced from the electrolysis of molten \( Al_2O_3 \), we will use Faraday's laws of electrolysis. Here’s a step-by-step solution:
### Step 1: Calculate the total charge (Q) passed through the electrolyte
The total charge can be calculated using the formula:
\[
Q = I \times t
\]
where:
- \( I \) is the current in amperes,
- \( t \) is the time in seconds.
Given:
- \( I = 4.0 \times 10^4 \) A,
- \( t = 6 \) hours = \( 6 \times 3600 \) seconds.
Calculating \( t \):
\[
t = 6 \times 3600 = 21600 \text{ seconds}
\]
Now, substituting the values into the charge formula:
\[
Q = 4.0 \times 10^4 \, \text{A} \times 21600 \, \text{s} = 864000000 \, \text{C} = 8.64 \times 10^8 \, \text{C}
\]
### Step 2: 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 \( F = 96500 \, \text{C/mol} \).
Substituting the values:
\[
n = \frac{8.64 \times 10^8 \, \text{C}}{96500 \, \text{C/mol}} \approx 8966.3 \, \text{mol}
\]
### Step 3: Determine the number of moles of aluminum produced
From the electrolysis of \( Al_2O_3 \), each mole of aluminum requires 3 moles of electrons (since \( Al^{3+} \) is reduced to \( Al \)). Therefore, the number of moles of aluminum produced is:
\[
\text{Moles of } Al = \frac{n}{3} = \frac{8966.3}{3} \approx 2988.77 \, \text{mol}
\]
### Step 4: 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 atomic mass of aluminum is \( 27 \, \text{g/mol} \):
\[
\text{Mass} = 2988.77 \, \text{mol} \times 27 \, \text{g/mol} \approx 80600.77 \, \text{g}
\]
### Final Result
The mass of aluminum produced is approximately:
\[
\text{Mass} \approx 8.06 \times 10^4 \, \text{g} \text{ or } 80600.77 \, \text{g}
\]
