Moles of Potassium Iodide Required to Form Complex P

To determine the number of moles of potassium iodide required to produce two moles of complex $\mathbf{P}$, we analyze the chemical reaction between potassium ferricyanide and potassium iodide.

1. Identification of the Reaction and Complex $\mathbf{P}$: When potassium iodide ($\text{KI}$) reacts with potassium ferricyanide ($\text{K}_3[\text{Fe(CN)}_6]$), a redox reaction occurs where iodide ions ($\text{I}^-$) act as a reducing agent to reduce iron from the $+3$ oxidation state in ferricyanide to the $+2$ oxidation state in ferrocyanide. Simultaneously, iodide is oxidized to iodine ($\text{I}_2$).

The complex $\mathbf{P}$ formed in this reaction is potassium ferrocyanide, $\text{K}_4[\text{Fe(CN)}_6]$.

2. Balanced Chemical Reaction: The balanced chemical equation for the reaction in aqueous medium is: $2\text{K}_3[\text{Fe(CN)}_6] + 2\text{KI} \rightleftharpoons 2\text{K}_4[\text{Fe(CN)}_6] + \text{I}_2$

In ionic form: $2[\text{Fe(CN)}_6]^{3-} + 2\text{I}^- \rightleftharpoons 2[\text{Fe(CN)}_6]^{4-} + \text{I}_2$

3. Stoichiometric Ratio: From the balanced equation:

• $2 \text{ moles of } \text{KI}$ are required to produce $2 \text{ moles of } \text{K}_4[\text{Fe(CN)}_6] \text{ (complex } \mathbf{P}\text{)}$.

Thus, the number of moles of potassium iodide required to produce two moles of complex $\mathbf{P}$ is 2.