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) Required electricity in Faraday to obtianed 20.0 g calcium from the molten `CaCl_(2)`
Ionization reaction: `CaCl_(2) to Ca^(2+) +2Cl^(-)`
Reduction reaction on cathode : `Ca^(2+)+2e^(-) to Ca`
`therefore 2" mol "e^(-) to 1" mol "Ca`
`therefore 2F to 40g` Ca
`therefore`Faraday required to produce 20 g Ca
`=(20gxx2F)/(40g)=1F`
(ii) Required electricity in Faraday to obtained 40.0 g aluminium from the molten `Al_(2)O_(3)`.
The following reaction occurs on cathode of molten `Al_(2)O_(3)`.
`Al_(2)O_(3(l))to 2Al^(3+)+3O^(2-)`
and reduction `Al^(3+)+3e^(-) to Al`
So, we get 1 mol Al from 3 mole `e^(-)`
`therefore` We get 27.0 g Al from 3F electricity.
So, the electricity required to get 40.0 g Al
`=(3Fxx40.0g)/(27.0g)=4.444F`