How much electricity in terms of Faraday is required to produce:
$(i)$ $20.0 \, g$ of $Ca$ from molten $CaCl_2$?
$(ii)$ $40.0 \, g$ of $Al$ from molten $Al_2O_3$?

(N/A) $(i)$ The reduction reaction for $Ca^{2+}$ is:
$Ca^{2+} + 2e^- \to Ca$
The molar mass of $Ca$ is $40 \, g/mol$.
To produce $1 \, mol$ $(40 \, g)$ of $Ca$,$2 \, F$ of electricity is required.
Therefore,electricity required to produce $20.0 \, g$ of $Ca = \frac{2 \, F}{40 \, g} \times 20.0 \, g = 1.0 \, F$.
$(ii)$ The reduction reaction for $Al^{3+}$ is:
$Al^{3+} + 3e^- \to Al$
The molar mass of $Al$ is $27 \, g/mol$.
To produce $1 \, mol$ $(27 \, g)$ of $Al$,$3 \, F$ of electricity is required.
Therefore,electricity required to produce $40.0 \, g$ of $Al = \frac{3 \, F}{27 \, g} \times 40.0 \, g = 4.44 \, F$.