How much electricity in terms of Faraday is required to produce `40.0 g` of `Al` from molter `Al_(2)O_(3)`?

Explanation

`Al^(+3) + 3ee^(-1) to Al`
1 mole of Al = 27g Al require electricity = 3F
40 g of Al require electricity
`=(3 xx 40)/27 = 4.44 F`