Aluminium oxide may be electrolysed at $$1000^\circ$$C to furnish aluminium metal (Atomic mass $$= 27$$ amu; $$1$$ Faraday $$= 96{,}500$$ Coulombs). The cathode reaction is $$\text{Al}^{3+} + 3e^- \longrightarrow \text{Al}^\circ$$. To prepare $$5.12$$ kg of aluminium metal by this method would require

Mass of aluminium to be produced: $$m = 5.12\ \text{kg} = 5120\ \text{g}$$

Molar mass of aluminium: $$M = 27\ \text{g mol}^{-1}$$

Moles of aluminium required: $$n = \frac{m}{M} = \frac{5120}{27}\ \text{mol} \approx 189.63\ \text{mol}$$

The cathode reaction is $$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$$. Each mole of $$\text{Al}^{3+}$$ needs $$3$$ moles of electrons, i.e. $$3$$ Faradays.

Total charge needed: $$Q = n \times 3 \times F$$ where $$F = 96\,500\ \text{C mol}^{-1}$$ is the Faraday constant.

Substituting, $$Q = 189.63 \times 3 \times 96\,500$$ $$Q \approx 568.89 \times 96\,500$$ $$Q \approx 5.49 \times 10^{7}\ \text{C}$$

Therefore, the charge required is $$5.49 \times 10^{7}$$ coulombs.

Option A which is: $$5.49 \times 10^{7}\ \text{C}$$