How many faradays of electricity are required to produce $13$ gram of aluminium from aluminium chloride solution? (Given: Molar mass of $Al=27.0$ gram $mo{l}^{-1}$).

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

The number of moles of $Al=\frac{13g}{27g/mol}=0.48mol$

1 mole of Al is produced by 3 moles of electrons.

0.48 mole of Al is produced by $3\times 0.48=1.44$ moles of electrons. This corresponds to $1.44$ faradays of electricity.

Note: 1 faraday of electricity corresponds to 1 mole of electrons.