The number of moles of electrons required to deposit 36 ~g of Al from an aqueous solution of Al ( NO ₃ )₃ is (At. wt. of Al =27 )

Step-by-step solution

Al ³⁺+ 3 ~mol 3 e⁻ 1 ~mol =27 ~g Al 27 ~g of Al is deposited by 3 moles of electrons 36 ~g Al will be deposited by electrons = 3 27 36=4 ~mol