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` ?

Explanation

In general for the reaction,
`M^(n+) + "ne"^(-) to M`
n faraday of electricity is required for reduction of one mole of `M^(n+)` to give M.
(i) `Ca^(2+) + 2e^(-) to Ca`
Thus, 1 mol of Ca, i.e, 40 g of Ca require electricity = 2F
`therefore 20 g` of Ca will require electricity = 1F
(ii) `Al^(3+) + 3e^(-) to Al`
Thus, 1 mol of Al, i.e. 27 g of Al require electricity = 3F
`therefore 40 g` of Al will require electricity `=3/27 xx 40 = 4.44 F`