Question

If during the electrolysis of molten $Al _{2} O _{3}$, the volume of $O _{2}( g )$ obtained at anode is $134.4\, L$ at STP, then the moles of $Al$ metal deposited at cathode in the same time, will be...........

Answer 1

Equivalents of $O _{2}( g )$ produce $\frac{134.4}{5.6}=24$
( $5.6\, L$ is the equivalent volume of $O _{2}$ gas)
$\therefore $ Equivalents of $Al$ metal deposited at cathode
$=24$
$\therefore $ Mass of $Al$ deposited at cathode
$=24 \times 9\, g =216\, g$
$\therefore $ Mol of $Al$ metal deposited at cathode
$=\frac{24 \times 9}{27}=8$