Question

If 10 g of $V_2{O}_5$ is dissolved in acid and is reduced to $V^{2+}$ by zinc metal, how many moles $I_2$ could be reduced by the resulting solution if it is further oxidised to $V{O}^{2+}$ ions?

[Assume no change in state of $Z{n}^{2+}$ ions] $(V=51,O=16,I=127)$

• A. $0.11moleof{I}_2$
• B. $0.22moleof{I}_2$
• C. $0.055moleof{I}_2$
• D. $0.44moleof{I}_2$

Answer 1

The correct option is A $0.11moleof{I}_2$

Given, $10g$ of $V_2{O}_5$

$\therefore$ No. of moles of $V_2{O}_5=\frac{10}{182}=0.0549$ moles $⟶\left(1\right)$

The reactions involved are:-

(i) $V_2{O}_5+10H^{+}+6e^{-}⟶2V^{2+}+5H_{2}O$

(ii) $V^{2+}+I_2+H_{2}O⟶2I^{-}+V{O}^{2+}+2H^{+}$

Here, it can be clearly seen that

$1$ mole $V_2{O}_5$ gives $2$ moles $V^{2+}$

& $1$ mole $V^{2+}$ can reduce $1$ mole $I_2$

Now, No. of moles of $V_2{O}_5=0.0549$ moles [$\because \left(1\right)$]

$\Rightarrow$ No. of moles of $V^{2+}=0.0549\times 2=0.1098$moles

$\therefore 0.1098$ moles $\approx 0.11$ moles of $V^{2+}$ can reduce $0.11$ moles of $I_2$.

So, the correct option is $A$