Calculate the mass of aluminium deposited from an aqueous solution of aluminum sulfate when a current of 3.0 A is passed for 2 hours. (Molar mass of Al = 27 g/mol; Faraday constant F = 96500 C/mol)

Aluminium ions are $Al^{3+}$.

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

This reaction shows that 3 moles of electrons are required to deposit 1 mole of aluminium.

One Faraday ($F$) is the charge of one mole of electrons, given as $96500 \text{ C}$.

$$\text{Moles of } e^- = \frac{Q}{F} = \frac{21600 \text{ C}}{96500 \text{ C/mol}} \approx 0.22383 \text{ mol } e^-$$

$$\text{Mass of } Al = \text{Moles of } Al \times \text{Molar mass of } Al$$ $$\text{Mass of } Al = 0.07461 \text{ mol} \times 27 \text{ g/mol} \approx 2.0145 \text{ g}$$

The calculated mass is approximately $2.01 \text{ g}$, which is closest to $2.02 \text{ g}$.

The final answer is $\boxed{\text{D } 2.02 \text{ g}}$.

6. In electrolytic purification processes, the impure metal to be purified is used as the anode. At the anode, the impure metal is oxidized to form metal ions, which then migrate to the cathode to be deposited as pure metal.

The final answer is $\boxed{\text{A anode}}$.

7. One faraday is equal to the charge of one mole of electrons. Numerically, this value is approximately $96500 \text{ coulombs}$.

The final answer is $\boxed{\text{B } 96 \text{ 500 coulombs}}$.