$2K\left(s\right)+2H_{2}O\left(l\right)\to 2KOH\left(aq\right)+H_2\left(g\right)$
If $3.0$ moles of potassium react with excess water, what volume of hydrogen gas will be produced?

The correct option is D $33.6L$
Given, $2K\left(s\right)+2H_{2}O\left(l\right)\to 2KOH\left(aq\right)+H_2\left(g\right)$

$2moles$ of Potassium on reaction gives $1mole$ of Hydrogen [ 22.4 at STP]

$2moles$ of $K\to 22.4Litres$ of $H_2$

$2moles$ of $K=\frac{3\times 22.4}{2}=33.6Litres$ of $H_2$

So, $33.6Litres$ of Hydrogen are produced.