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) `Cu^(2+)+2e^(-)toCa`
Thus, 1 mol of Ca, i.e., 40 g of Ca require electricity=2F`therefore20` g of Ca will require electricity=1F
(ii) `Al^(3+)+3e^(-)toAl` . Thus 1 mol of Al, i.e., 27g of Al required electricity=3F
`therefore40g` of Al will require electricity=`(3)/(27)xx40=4.44E`