Question

In an experiment, $20$ g of vanadium ($V$) oxide (molar mass$=182$ g/mol) was reduced by excess of zinc dust in acidic solution to vanadium (II) ions. If the required number of moles of iodine to reoxidise vanadium (II) to $V{O}^{2+}$ is $22\times {10}^{-x}$, then what is the value of $x$?

Answer 1

According to given conditions,
$V_2{O}_5⟶V^{2+}⟶{VO}^{2+}$
Given,
$\underset{20g0.11mol}{V_2{O}_5}⟶\underset{0.22mol}{2V^{2+}}$
$\underset{+2}{2V^{2+}}+\underset{0}{2I_2}⟶\underset{+4}{{2VO}^{2+}}+\underset{-2}{4I^{-}}$
Thus, $\underset{0.11mol}{V_2{O}_5}\equiv 2V^{2+}\equiv 2{VO}^{2+}\equiv \underset{0.22mol}{2I_2}$
So, required number of moles of iodine to reoxidise is $0.22$ mol.