Question

$KI$ reacts with $H_{2}S{O}_4$ producing $I_2$ and $H_{2}S$. The volume of $0.2M{H}_{2}S{O}_4$ required to produce $0.1$ mol of $H_{2}S$ is:

• A. $4L$
• B. $2.5L$
• C. $3.8L$
• D. $5L$

Answer 1

The correct option is B $2.5L$

We have

$8KI+5H_{2}S{O}_4\to 4K_{2}S{O}_4+H_{2}S+4I_2+4H_{2}O$

From the reaction above:

5 moles of $H_{2}S{O}_4$ is producing 1 mole of $H_{2}S$

So, 0.1 mole of $H_{2}S$ is produced by 0.5 mole of $H_{2}S{O}_4$

Moles of $H_{2}S{O}_4$ = 0.5 mol

Now

Molarity = $\frac{moles}{volume}=\frac{0.5}{Volume}$

$0.2=\frac{0.5}{volume}$

$\therefore$ Volume = 2.5 L