The number of electrons required to reduce 4.5 10⁻⁵ ~g of Al is

Step-by-step solution

27 Al ³⁺ +3 e⁻ 27 ~g Al 27 ~g of Al is reduced by =3 6.023 10²³ e⁻ s 4.5 10⁻⁵ ~g of Al will be reduced by array l = 3 6.023 10²³ 4.5 10⁻⁵ 27 =3.01 10¹⁸ electrons array