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Q.

The number of moles of KI required to produce 0.4 mole K₂ Hg I₄ is 4KI + HgCl₂ → K₂HgI₄ + 2KCl

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a

1

b

16

c

1.6

d

3

answer is D.

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Detailed Solution

\large \mathop {4KI}\limits_{'X'\;mole}^{4\;mole} + \;HgC{l_2}\; \to \;\mathop {{K_2}Hg{I_4}}\limits_{0.4\;mole}^{1\;mole} \; + \;2KCl

\large X\; = \;\frac{{0.4 \times 4}}{1}\; = \;1.6

$X$ $mole$ $4$ $mole$ $$ $4$ $KI$ $+$ $HgCl$ $2$ $$ $\to$ $0.4$ $mole$ $1$ $mole$ $$ $K$ $2$ $$ $HgI$ $4$ $$ $+$ $2$ $KCl$ $X = \frac{0.4 \times 4}{1} = 1.6$ $X$ $=$ $10.4$ $\times$ $4$ $$ $=$ $1.6$

This shows the stoichiometric calculation where 4 moles of KI react with 1 mole of HgCl₂ to form K₂HgI₄ and 2KCl.
The calculation determines the value of $X$ $X$ , giving 1.6 moles.

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