Question

$KI+I_{2}HN{O}_3\to HI{O}_3+KI{O}_3+N{O}_2$
If 3 moles of KI and 2 moles of $I_2$ are mixed with excess $HN{O}_3$, then volume of $N{O}_2$ gas evolved at NTP is:

• A. 716.8 litre
• B. 1075.2 litre
• C. 44.8 litre
• D. 67.2 litre

Answer 1

Answer: (C)

44.8 litre

according to the equation,

1 mole of $KI$ is reacting with 1 mole of $I_2$

so, for 2 mole of $I_2$ we require 2 mole of KI

so, $I_2$ is the limiting reagent

1 mole of $I_2$ is producing 1 mole of $N{O}_2$

so, 2 mole of $I_2$ produce 2 moles of $N{O}_2$

1 mole of gas at NTP = 22.4 L

so, volume of 2 mole of $N{O}_2$ = $2\ast 22.4$

= $44.8L$