How muCHM 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)`

Answer

`a. 1F`,`b. 4.44F`

Explanation

`a.` `Ca^(2+) +2e^(-) rarr Ca`
Thus, `1 mol of Ca, i.e., 40 g ` of `Ca` require electricity `=2F`
`:. 20 g ` of `Ca ` will require electricity `=3F`
`b.` `Al^(3+)+3e^(-) rarr Al`
`:. 40 g of Al` will require electricity `=(3)/(27)xx40`
`=4.44F`