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` ml of `I_(2)`
D. `0.44` ml of `I_(2)`

Answer

A

Explanation

`V_(2)O_(5) +10H^(+) +6e^(-) rarr 2V^(2+) +5H_(2)O`
n-factor for `V_(2)O_(5) = 6`
`V^(2+) rarr VO^(2+) +2e` (n-factor =2)
`I_(2) +2e rarr 2I^(-) , V^(2+) +I_(2) rarr VO^(2+) +2I^(-)`
moles of `I_(2) =` moles of `V^(2+) =2` moles of `V_(2)O_(5)`
`= 2xx (10)/(182) = 0.109 ~~ 0.11`