The equivalent weight of a substance is defined as its molecular weight divided by the number of electrons involved in the reaction (n-factor). For KCN, the molecular weight is $M$. In the reaction with ferric hydroxide, the chemical equation is: $$6 \mathrm{KCN} + \mathrm{Fe}(\mathrm{OH})_{3} \rightarrow \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] + 3 \mathrm{KOH}$$ In this reaction, 6 moles of KCN react with 1 mole of $\mathrm{Fe}(\mathrm{OH})_{3}$. The cyanide ions ($\mathrm{CN}^-$) coordinate with the iron center. Considering the stoichiometry, the n-factor for KCN in this specific complex formation is $0.5$ (since 6 moles of KCN are required to form the complex containing 6 cyanide ligands, effectively distributing the reaction capacity). Thus, the equivalent weight $E$ is: $$E = \frac{M}{0.5} = 2M$$ This demonstrates that the equivalent weight of KCN can be twice its molecular weight in this reaction. )%2013.png?alt=media&token=7ce9c8e8-454d-4a6d-bb6d-e129dffba401" alt="">