Calculate the number of coulombs required to deposit 40.5 g of Al when the electrode reaction is ,
`Al^(3+)+3e^(-)rarr Al`

Answer

`4.342xx10^(5) C`

Explanation

`Al^(3+)(aq)+underset(3xx96500C)(3e^(-)) to underset(27 g)(Al(s))`
To deposit 27 g of Al(s), the amount of charge required `=3xx96500" C"`
To deposit, 40.5 g of Al (s), the amount of charge required`=((3xx96500C))/((27 g))xx(40.5 g)=4.342xx10^(5)C`.