Calculate the mass of aluminum deposited from molten aluminum oxide when a current of 40,000 A is passed for 1 hour.

The current ($I$) is $40,000 \text{ A}$ and the time ($t$) is $1 \text{ hour}$.
First, convert time to seconds:

$$t = 1 \text{ hour} \times 60 \text{ min/hour} \times 60 \text{ s/min} = 3600 \text{ s}$$

Now, calculate the charge:

$$Q = I \times t$$ $$Q = 40,000 \text{ A} \times 3600 \text{ s}$$ $$Q = 144,000,000 \text{ C}$$

Using Faraday's constant ($F = 96,500 \text{ C/mol e⁻}$):

$$n_e = \frac{Q}{F}$$ $$n_e = \frac{144,000,000 \text{ C}}{96,500 \text{ C/mol e⁻}}$$ $$n_e \approx 1492.228 \text{ mol e⁻}$$

Aluminum is produced from $\text{Al}^{3+}$ ions, which gain electrons to form solid aluminum:

$$\text{Al}^{3+} + 3\text{e}^- \to \text{Al}$$

From this reaction, 3 moles of electrons are required to produce 1 mole of aluminum.

$$n_{\text{Al}} = \frac{n_e}{3}$$ $$n_{\text{Al}} = \frac{1492.228 \text{ mol e⁻}}{3 \text{ mol e⁻/mol Al}}$$ $$n_{\text{Al}} \approx 497.409 \text{ mol Al}$$

The molar mass of aluminum ($\text{Al}$) is $27 \text{ g/mol}$.

$$m_{\text{Al}} = n_{\text{Al}} \times M_{\text{Al}}$$ $$m_{\text{Al}} = 497.409 \text{ mol} \times 27 \text{ g/mol}$$ $$m_{\text{Al}} \approx 13430.043 \text{ g}$$

$$m_{\text{Al}} = \frac{13430.043 \text{ g}}{1000 \text{ g/kg}}$$ $$m_{\text{Al}} \approx 13.43 \text{ kg}$$

The mass of aluminum produced is $\text{13.4 kg}$.