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

(a) `Ca^(2+)+2e^(-) to Ca` ,
1 mol Ca is produced from 2 mole or 40 g Ca is produced from 2 F .
` :. ` 20 g of Ca is produced from 1 F of electricity .
Thus , 1 faraday of electricity will be required to produce `20*0` g of Ca from molten `CaCl_(2)` . ,brgt (b) `al^(3+)+3e^(-) to Al`
27 g Al is produced from 3 F
` :. " 20 g of Al requires " = 3/27 xx 40 = 4 * 44 ` F
`4*44 ` F of electricity .
Hence, `4*44` faraday of electricity will be required to produced `40*0` g Al from molten `Al_(2)O_(3)`