Q.

If 10 g of V2O5 is dissolved in acid and is reduced to V2+ by zinc metal, how mole I2 could be reduced by the resulting solution if it is further oxidised to VO2+ ions ?[Assume no change in state of Zn2+ ions] (V = 51, O = 16, I = 127):

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a

0.11 mole of I2

b

0.22 mole of I2

c

0.055 mole of I2

d

0.44 mole of I2

answer is A.

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Detailed Solution

Mole of V2O5 = 1051 × 2 + 5 × 16 = 10102 + 80                       = 10182 =.055 Mole of V+2 = .055 × 2   = .1098 mole ≃ 0.11 V+2 → V+4O+2  + 2e- ⇒  Moles  of I2 = Moles of V+2 = .11
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