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

Explanation

`Ca^(2+)to2e^(-)toCa`
thus, 1 mol of Ca, i.e., 40 of Ca require=2F
electricity
`therefore20`g of Ca requrie=1F of electricity
(ii). `Al^(3+)+3e^(-)toAl`
Thus, 1 mole of Al, i.e., 27g of Al require=
3F electricity
`therefore40g` of Al will require electricity
`=(3)/(27)xx40=4.44` F of electricity.