How much electricity in terms of Faraday is required to produce.
`a.` `20.0g` fo `Ca` from molten `CaCl_(2)`
`b.` `40g` of `Al` from molten `Al_(2)O_(3)`

Explanation

(i) `" " Ca^(2+)(aq)+underset(2F)2e^(-) to underset(40g)(Ca(s))`
To produce 40 g of `CaCl_(2)`, charge needed =2F
To produce 20 g of `CaCl_(2)`, charge needed=1F
(ii) `Al_(2)O_(3)+underset(6F)6e^(-)to underset((2xx27)g)(2Al)+3O^(2-)`
To produce 54 g of Al, charge needed =6F
To produce 40 g of Al, charge needed `=((40g))/((54g))xx(6F)=4.44F`.