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 of $I_2$ could be reduced by the resulting solution if it is further oxidized to $V{O}^{2+}$ ions? (Atomic mass of $V$ is $51$ g/mol)

• A. $0.11$ mol of $I_2$
• B. $0.22$ mol of $I_2$
• C. $0.055$ mol of $I_2$
• D. $0.44$ mol of $I_2$

Answer 1

Answer: (A)

$0.11$ mol of $I_2$

The reaction is as follows:

$V_2{O}_5+10H^{⨁}+6e^{-}\to 2V^{2+}+5H_{2}O$

Milliequivalents of $V_2{O}_5=\frac{10\times 1000}{\frac{192}{6}}$

Millimoles of $V_2{O}_5=\frac{10\times 1000}{192}$

Millimoles of $V^{2+}=\frac{10}{192}\times 1000\times 2$
$(V^{2+}\to V{O}^{2+}+2e^{-}\Rightarrow n$-factor $=2$)

Milliequivalents of $V^{2+}$ (against $I_2$) $=\frac{20}{192}\times 1000\times 2=$ milliequivalents of $I_2$

Millimoles of $I_2=\frac{\left(\frac{20}{192}\times 1000\times 2\right)}{2}=104$

Moles of $I_2=0.104$ $[\because I_2+2e^{-}\to 2I^{⊝}]$