Question

If during the electrolysis of molten $A{l}_2{O}_3$, the volume of $O_2(g)$ obtained at anode is $134.4L$ at STP, then the moles of $Al$ metal deposited at cathode in the same time, will be...........

Answer 1

Answer: 8

Equivalents of $O_2(g)$ produce $\frac{134.4}{5.6}=24$
( $5.6L$ is the equivalent volume of $O_2$ gas)
$\therefore$ Equivalents of $Al$ metal deposited at cathode
$=24$
$\therefore$ Mass of $Al$ deposited at cathode
$=24\times 9g=216g$
$\therefore$ Mol of $Al$ metal deposited at cathode
$=\frac{24\times 9}{27}=8$