Chemistry - Electrochemistry Question with Solution | TestHub

$\underset{27}{{\mathrm{Al}}^{3+}}+3e^{-}⟶\underset{27\mathrm{g}}{\mathrm{Al}}$ $27\mathrm{g}$ of $\mathrm{Al}$ is reduced $\mathrm{by}=3\times 6.023\times 10^{23}{e}^{-}\mathrm{s}$ $4.5\times 10^{-5}\mathrm{g}$ of Al will be reduced by $$\begin{array}{l}=\frac{3\times 6.023\times 10^{23}\times 4.5\times 10^{-5}}{27}\\ =3.01\times 10^{18}\text{ electrons }\end{array}$$