(i) How much electricity in terms of Faraday required to produce.

20.0 g of Ca from molten $CaC{l}_{2}P$

(ii) How much electricity in terms of Faraday required to produce.

(i) Electrode reaction : $C{a}^{2+}\left(aq\right)+\underset{2F}{2e^{-}}\to \underset{40g}{Ca\left(s\right)}$

To produce 40g of Ca from molten $CaC{l}_2$, electricity required = 2F To produce 20 g Ca from $CaC{l}_2,$ electricity required = 1F

(ii) Electrode reaction $A{l}_2{O}_3\left(l\right)+\underset{6F}{6e^{-}}\to \underset{\left[\underset{=54g}{(2+3O\times 27)}\right]}{2Al}+3O^{2-}$

To produce 54g Al from molten $A_2{O}_3$, electricity required = 6F To produce 40 g Al from molten $A{l}_2{O}_3$, electricity required

$=\frac{\left(6F\right)\times \left(40g\right)}{\left(54g\right)}=4.44F$