Question

The number of moles of KI required to produce 0.1 moles of ${{K}_{2}}[Hg{{I}_{4}}]$is?
A. 1.6
B. 0.8
C. 3.2
D. 0.4

Answer 1

-To produce ${{K}_{2}}[Hg{{I}_{4}}]$from KI we need a chemical. The required chemical is $HgC{{l}_{2}}$ (mercuric chloride).
-The name of the KI is potassium Iodide.
-The name of the${{K}_{2}}[Hg{{I}_{4}}]$is potassium mercuric iodide.
-The colour of the potassium mercuric iodide is yellow.

Complete step by step solution:
-The formation of ${{K}_{2}}[Hg{{I}_{4}}]$ from KI is a two-step process.
Step-1: Potassium iodide reacts with $HgC{{l}_{2}}$ (Mercuric chloride) and forms $Hg{{I}_{2}}$ (Mercuric Iodide).
$2KI+HgC{{l}_{2}}\to Hg{{I}_{2}}+2KCl$
Step-2: The formed $Hg{{I}_{2}}$ (Mercuric Iodide) reacts with excess amount of KI and forms potassium mercuric iodide (${{K}_{2}}[Hg{{I}_{4}}]$).
$Hg{{I}_{2}}+2KI\to \underset{\text{potassium mercuric iodide}}{\mathop{{{K}_{2}}[Hg{{I}_{4}}]}}\,$
-The overall reaction can be written as follows.
$4KI+HgC{{l}_{2}}\to {{K}_{2}}[Hg{{I}_{4}}]+2KCl$
-From the above equation we can say that four moles of potassium iodide reacts with Mercuric iodide and forms one mole of${{K}_{2}}[Hg{{I}_{4}}]$(potassium mercuric iodide).
1 moles of ${{K}_{2}}[Hg{{I}_{4}}]$ requires 4 moles of KI
0.1 moles of ${{K}_{2}}[Hg{{I}_{4}}]$ requires = 0.1 (4) = 0.4 moles of KI.
-Therefore to produce 0.1 moles of ${{K}_{2}}[Hg{{I}_{4}}]$, there is a requirement of 0.4 moles of KI.

So, the correct option is D.

Additional information:
- Potassium mercuric iodide is a yellow colour solid and it is odourless.
-Potassium mercuric iodide is a salt and it is used as Nessler's reagent.
- Potassium mercuric iodide is soluble in water.
-The colour of the alkaline Potassium mercuric iodide is pale orange.

Note: Potassium mercuric iodide is used widely to determine the number of ammonium compounds. The structure of Potassium mercuric iodide is as follows.

0.09mol/L Potassium mercuric iodide in 0.25mol/L potassium hydroxide is called Nessler’s reagent.