The number of moles of electrons required to deposit $36 \mathrm{~g}$ of $\mathrm{Al}$ from an aqueous…
The number of moles of electrons required to deposit $36 \mathrm{~g}$ of $\mathrm{Al}$ from an aqueous solution of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is (At. wt. of $\mathrm{Al}=27$ )
$4$
$2$
$3$
$1$
Solution
$\mathrm{Al}^{3+}+\underset{3 \mathrm{~mol}}{3 e^{-}} \longrightarrow \underset{1 \mathrm{~mol}=27 \mathrm{~g}}{\mathrm{Al}}$
$\because 27 \mathrm{~g}$ of $\mathrm{Al}$ is deposited by 3 moles of electrons
$\therefore 36 \mathrm{~g} \mathrm{Al}$ will be deposited by electrons
$=\frac{3}{27} \times 36=4 \mathrm{~mol}$