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Stoichiometry

Number of mol...

Question

Number of moles of $K I$ required to prepare $0.4$ moles of $K_{2} H g I_{4}$ , when $H g C l_{2}$ reacts with $K I$ , is:

0.4

moles of

K I

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0.8

moles of

K I

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1.6

moles of

K I

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3.2

moles of

K I

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Solution

The correct option is D $1.6$ moles of $K I$
The reaction is $H g C l_{2} + 4 K I \to K_{2} H g I_{4} + 2 K C l$ .
$4$ moles of $K I$ are required to prepare $1$ mole of $K_{2} H g I_{4}$ .
Thus, to prepare $0.4$ moles of $K_{2} H g I_{4}$ , the number of moles of $K I$ required is $4 \times 0.4 = 1.6$ mol.

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