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_2O_5+10H^(o+)+6e^(-)to2V^(2+)+5H_2O`
(Mw of `V_2O_5=182," Eq of "V_2O_5=(182)/(6))`
`(10)/(182) " mol of "V_2O_5-=2xx(10)/(182) " mol of "V^(2+)`
(ii). `H_2O+V^(2+)+I_2to2I^(ɵ)+VO^(2+)+2H^(o+)`
`(20)/(182)=(10)/(91) " mol of "V^(2+)=(10)/(91) " mol of "I_2`
`" mol of "I_2=(10)/(91)`