The number of moles of KI required to prepare one mole of K₂[HgI₄] is

The formation of ${\mathrm{K}}_2\left[\mathrm{Hg}{\mathrm{I}}_4\right]$ from KI is a two-step process.
Step-1: Potassium iodide reacts withHgCl₂(Mercuric chloride) and forms HgI₂(Mercuric Iodide).
2KI+HgCl₂→HgI₂+2KCl
Step-2: The formedHgI₂(Mercuric Iodide) reacts with excess amount of KI and forms potassium mercuric iodide (K₂[HgI₄]).
HgI₂+2KI→K₂[HgI₄] potassium mercuric iodide
-The overall reaction can be written as follows.
4KI+HgCl₂→ K₂[HgI₄]+2KCl
From the above equation we can say that four moles of potassium iodide reacts with Mercuric iodide and forms one mole ofK₂[HgI₄](potassium mercuric iodide).