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 \[{{I}_{2}}\]
Correct Answer: A
Solution :
[a] \[\underset{+5}{\mathop{{{V}_{2}}{{O}_{5}}}}\,+6{{e}^{-}}\to \underset{+2}{\mathop{2{{V}^{2+}}}}\,\] Molecular mass of \[{{V}_{2}}{{O}_{5}}=51\times 2+5\times 16=182\] 1 mol of \[{{V}_{2}}{{O}_{5}}\]produces 2 moles of \[{{V}^{2+}}\] \[\therefore \frac{10}{182}\]moles of \[{{V}_{2}}{{O}_{5}}\]produce \[2\times \frac{10}{182}=0.11\]moles of \[{{V}^{2+}}\] \[\underset{+2}{\mathop{{{V}^{2+}}}}\,\to V{{\underset{0}{\mathop{O}}\,}^{2+}}+2{{e}^{-}}\] (1) \[\underset{0}{\mathop{{{I}_{2}}}}\,+2{{e}^{-}}\to \underset{-1}{\mathop{2{{I}^{-}}}}\,\] (2) On adding (1) and (2) \[{{V}^{2+}}+{{I}_{2}}\to 2{{I}^{-}}+V{{O}^{2+}}\] 1 mole of \[{{V}^{2+}}\]reduces 1 mole of \[{{I}_{2}}\] \[\therefore 0.11\]moles of \[{{V}^{2+}}\]reduce 0.11 moles of \[{{I}_{2}}\]
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