Find the volume of `Cl_(2)` at `NTP` produced during electrolysis of `MgCl_(2)` which produces `6.6 g Mg`. `("At.wt. of " Mg = 24.3)`

Explanation

At cathode `Ma^(2+)+2erarrMg`
At anode `2Cl^(-)rarrCl_(2)+2e`
`because` Equivalent of Mg at cathod = Equivalent of `Cl_(2)`
at anode `therefore(6.6)/(24.3//2)=(w_(Cl_(2)))/(35.5)rArrw_(Cl_(2))=19.28g`
Now at NTP PV `=(w)/(m)RT`
`1xxV=(19.28xx0.0821xx273)/(71)`
`rArrV=6.08"litre"~~6"litre"`