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` mole of `I_(2)`
B. `0.22` mole of `I_(2)`
C. `0.055` mole of `I_(2)`
D. `0.44` mole of `I_(2)`

Answer

A

Explanation

Mole of `V_(2)O_(5)=(10)/(51xx2+5xx16)=(10)/(102+80)`
`(10)(182)=.055`
Mole of `V^(+2)=.055xx2`
`=.1098 mol e ~=0.11`
`V^(+2) rarr overset(+4)(VO^(+2))+2e`
`implies Mol es of I_(2)=Mol es of V^(+2)=.11`