$2 K I + I_{2} + 22 H N O_{3} \to 2 H I O_{3} + 2 K I O_{3} + 22 N O_{2}$
If 3 mole of $K l$ & 2 moles $I_{2}$ , are reacted with excess of $H N O_{3}$ .
What is the volume of $N O_{2}$ gas evolved at NTP?

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Solution

$2 K I + I_{2} + 2 H N O_{3} \to 2 H I O_{3} + 2 K I O_{3} + 22 N O_{2} 2 m o l e s o f K I = 1 m o l e o f I_{2} = 22 m o l e s o f N O_{2} 1 m o l e o f K I = \frac{1}{2} m o l e o f I_{2} = 11 m o l e s o f N O_{2} 3 m o l e s o f K I = \frac{3}{2} m o l e o f I_{2} = 33 m o l e s o f N O_{2} H e n c e, t h e n u m b e r o f m o l e s o f N O_{2} e v o l v e d i s = 33 m o l e s 1 m o l e o f g a s o c c u p i e s 22.4 l a t N . T . P 33 m o l e s o f N O_{2} o c c u p i e s = 33 \times 22.4 l a t N . T . P = 739.2 l a t N . T . P$

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