In the reaction, $VO+{Fe}_2{O}_3\to FeO+V_2{O}_5$, the eq.wt.of $V_2{O}_5$ is equal to its :

The correct option is D molecular weight/6

$VO+{Fe}_2{O}_3⟶FeO+V_2{O}_5$

Oxidation no. of $V$ in $VO=+2$

Oxidation no. of $V$ in $V_2{O}_5=+5$

Valence factor $=$ no. of atoms $\times$ change in oxidation no.

$\therefore$ Valence factor for $V_2{O}_5=2\times |5-2|=6$

As we know that,

$\text{eq. wt.}=\frac{\text{mol. wt.}}{\text{valence factor}}$

$\therefore$ Equivalent weight of $V_2{O}_5$ for the given change $=\frac{\text{Molecular weight}}{6}$

Hence for the given change, the equivalent weight of $V_2{O}_5$ will be $\frac{\text{Molecular weight}}{6}$.