If 10.0 g `V_2O_5` is dissolbed in acid and reduced to `V^(2+)` by treatment with tin (Sn) metal how many moles of `I_2` could be reduced by the resulting `V^(2+)` solution as it is oxidised to `V^(4+)`? (Atomic weight of V is 51)

Explanation

`V_(2)^(5+)+6erarr2V^(2+)`
`ZnrarrZn^(2+)+2e`
`V^(2+)rarrV^(4+)+2e`
`I_(2)+2erarr2I^(-)`
and Meq.of `V^(2+)(v.f=3)=` Meq.of ` V_(2)O_(5)(v.f=6)`
`=(10)/(182//6)xx1000`
Meq.of `V^(2+)(v.f.=2)=(10)/(182//6)xx1000xx(2)/(3)`
Meq.of `V^(2+)(v.f=2)=` Meq.of `I_(2)`
`(10xx6)/(182)xx1000xx(2)/(3)=` Meq.of `I_(2)`
or Meq.of `I_(2)=219.78`
`m` Mole of `I_(2)=(219.78)/(2)=109.89`
Mole of `I_(2)=(109.89)/(1000)`
Mole of `I_(2)=0.1098`