${\mathrm{Al}}_2{\mathrm{O}}_3$ isreduced by electrolysis at low potentials and high currents.if $4.0\times {10}^4$ amperes of current is passed through molten ${\mathrm{Al}}_2{\mathrm{O}}_3$ 6 hours, what mass of aluminum is produced  (Assume 100%current efficiency, at. mass of A1 $=27\mathrm{g}{\mathrm{mol}}^{-})$

Eq.wt.of

$\begin{array}{l}\mathrm{Al}=\text{ At wt of }\mathrm{Al}/\text{ Valency }=27/3=9.\\ \mathrm{Z}=\text{ Electrochemical equivalent }\\ =\text{ g. eq. }\mathrm{wt}/96500\text{ coulomb }\end{array}$

$\begin{array}{l}\therefore \text{ g. eq. wt. of }\mathrm{Al}=Z\times 96500\mathrm{C};9\mathrm{g}=Z\times 96500\mathrm{C}\\ \therefore Z=\frac{9g}{96500\mathrm{C}}.\end{array}$

We know

Mass of  $\mathrm{Al}=\mathrm{Z}\times I\times t$

mass of $\begin{array}{r}\mathrm{Al}=\frac{9g}{96500\mathrm{C}}\times 4\times {10}^4\mathrm{amp}\cdot \times 6\times 60\times 60\mathrm{s}\\ =8.1\times {10}^4\mathrm{g}\cdot \frac{\text{ amp.s }}{\mathrm{C}}=8.1\times {10}^4\mathrm{g}\end{array}$