How much electricity in terms of Faraday is required to produce : 40.0 g of Al from molten A1203?

Explanation

`underset(6F)(Al_(2)O_(3))+6e^(Theta)rarrunderset((2xx27)g)(2Al+3O^(2-))`
To produce 64 g of Al, charge needed = 6 F
To produce 40 g of Al, charge needed `=((40g))/((50g))xx(6f)=4.44f`