If 10 g of V₂O₅ is dissolved in acid and is reduced to V²⁺ by zinc metal, how many moles of iodine, could be reduced by the resulting solution if it is further oxidised to VO²⁺ ions ?
(V = 51, O = 16, I = 127) :

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0.44 mole of I₂

0.22 mole of I₂

0.11 mole of I₂

0.055 mole of I₂

answer is A.

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Detailed Solution

$Mole of V_{2} O_{5} = \frac{10}{51 \times 2 + 5 \times 16} = \frac{10}{102 + 80} = \frac{10}{182} = .055$

$\begin{array}{l} Mole of V^{+ 2} = 0.055 \times 2 \\ = 0.1098 mole ≃ 0.11 \\ V^{+ 2} ⟶ V O^{+ 2} + 2 e \\ I_{2} + 2 e ⟶ 2 I^{-} \\ \Rightarrow Moles of I_{2} = Moles of V^{+ 2} = 0.11 \end{array}$

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