Question

Equivalent weight of $Fe{C}_2{O}_4$ in the change $Fe{C}_2{O}_4\to F{e}^{3+}+2C{O}_2$, is:

• A. $M/3$
• B. $M/6$
• C. $M/2$
• D. $M/1$

Answer 1

The correct option is A $M/3$

${Fe}^{+2}\to {Fe}^{+3}+e^{-}$

${C_2{O}_4}^{-2}\to {2CO}_2+2e^{-}$

Adding both the reactions, we get,

${Fe}^{+2}{{+C}_2{O}_4}^{-2}\to {2CO}_2+{Fe}^{+3}+3e^{-}$

Equivalent weight is defined as the molecular weight divided by the number of moles of electron required by the molecule.

Hence, the equivalent weight is equal to $\frac{M}{3}$.

Therefore, option $A$ is correct.