Question

How many moles of $H_{2}S{O}_4$ are required to produce $1$ mol of $H_{2}S$ when $KI$ reacts with $H_{2}S{O}_4$ producing $I_2$ and $H_{2}S$?

• A. $5$
• B. $4$
• C. $3$
• D. $2$

Answer 1

Answer: (A)

$5$

The balanced reaction is as follows:
$8KI+\underset{\left(5mol\right)}{5H_{2}S{O}_4}\to 4K_{2}S{O}_4+4I_2+\underset{\left(1mol\right)}{H_{2}S}+4H_{2}O$
So, $5$ mol of $H_{2}S{O}_4$ is required to produce $1$ mol of $H_{2}S$ when $KI$ reacts with $H_{2}S{O}_4$ producing $I_2$ and $H_{2}S$.