If `10g` of `V_(2)O_(5)` is dissolved in acid and is reduced to `V^(2+)` by zinc metal, how many mole `I_(2)` could be reduced by the resulting solution if it is further oxidised to `VO^(2+)` ions? [Assume no change in state of `Zn^(2+)` ions] (`V=51`, `O=16`, `I=127`)

Options

A. 0.11 " mol of "`I_(2)`
B. `0.22 " mol of "I_(2)`
C. `0.055 " mol of "I_(2)`
D. `0.44 " mol of "I_(2)`

Answer

A

Explanation

(i). `V_(2)O_(5)+10H^(o+)+6e^(ɵ)to2V^(2+)+5H_(2)O`
`impliesmEq V_(2)O_(5)=(10xx1000)/((192)/(6))`
`mmol (V_(2)O_(5))=(10xx1000)/(192)`
`impliesmmol V^(2+)=(10)/(192)xx1000x2`
`(V^(2+)toVO^(2+)+2e^(-)implies"n-factor"=2)`
`mEq V^(2+) ("against" I_(2))=(20)/(192)xx1000xx2=mEq I_(2)`
`implies mmol I_(2)=((20)/(192)xx1000xx2))/(2)=104`
`[becauseI_(2)+2e^(-)to2I^(ɵ)]`