The number of moles of potassium iodide required to produce two moles of P is ______.

Potassium ferricyanide is $$K_3[Fe(CN)_6]$$, in which iron is in the $$+3$$ oxidation state. When iodide ion $$I^-$$ is added, a redox equilibrium is set up in which iron(III) is reduced to iron(II) and iodide is oxidised to iodine. The iron(II) product is potassium ferrocyanide $$K_4[Fe(CN)_6]$$, designated as the complex P.

Step 1 - Write the half-reactions
Reduction: $$[Fe(CN)_6]^{3-} + e^- \rightarrow [Fe(CN)_6]^{4-}$$
Oxidation: $$I^- \rightarrow \dfrac{1}{2}I_2 + e^-$$

Step 2 - Equalise electrons and add the half-reactions
Each half-reaction involves one electron, so they can be added directly:

$$[Fe(CN)_6]^{3-} + I^- \rightarrow [Fe(CN)_6]^{4-} + \dfrac{1}{2}I_2$$

Step 3 - Remove the fractional coefficient
Multiply the entire equation by 2:

$$2\,[Fe(CN)_6]^{3-} + 2\,I^- \rightarrow 2\,[Fe(CN)_6]^{4-} + I_2$$

Replacing the complex ions with their potassium salts gives:

$$2\,K_3[Fe(CN)_6] + 2\,KI \rightarrow 2\,K_4[Fe(CN)_6]\;(\text{= P}) + I_2$$

Step 4 - Determine the stoichiometry for potassium iodide
From the balanced equation, 1 mol of $$KI$$ is required to form 1 mol of $$K_4[Fe(CN)_6]$$. Therefore, to obtain 2 mol of $$K_4[Fe(CN)_6]$$ (complex P) we need:

$$2\;\text{mol P} \times \dfrac{1\;\text{mol }KI}{1\;\text{mol P}} = 2\;\text{mol }KI$$

Hence, the number of moles of potassium iodide required is 2.

Final answer: 2