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_(2)O_(3)`.

Explanation

(i) `Ca^(2+)+2e^(-)toCa`
1 mole of Ca i.e., 40 g require electricity = 2F
20 g of Ca will require electricity `=(2xx20)/(40)F`
=1F
(ii) `Al^(3+)+3e^(-)toAl`
1 mol of Al =27 g
`:.` 27 g of Al require electricity =3F
40 g of Al will require electricity `=(3)/(27)xx40F`
=4.44F.