How many grams of `Cl_(2)(g)` can be produced by the electrolysis molten `NaCl` with a current of `5.5A` for `25 m i n` ? `[` Atomic weight of `Cl=35.5 am u]`

Explanation

Molten `NaCloverset(El ectrolysis)rarr Na^(o+)+Cl^(c-)`
Number of Faradays `(Ixxt "in seconds" )/(96500C)`
`=(5.5s xx 25 xx 60s)/(96500C)=0.085F`
First method
Oxidation of `Cl^(c-)` at anode `:`
`2Cl^(c-) rarr Cl_(2)+2e^(-)`
`2e^(-) =2F=1 mol Cl_(2)=71g`
`-0.085F=(71)/(2) xx 0.085=3.017g Cl_(2)`
Second method
`1F=1` equivalent of `Cl_(2)=(71g)/(Charg e)=(71g)/(2)g`
`:. 0.085F=(71)/(2)xx0.085=3.017 g Cl_(2)`