If 30 mL of $H_2$ and 20 mL of $O_2$ reacts to form $H_{2}O$, what is left at the end of the reaction ?

The correct options are
A 10 mL of $H_2$
D 5 mL of $O_2$
$1ml$ of $H_2$ reacts with $0.5ml$ of $O_2$ to form $1ml$ of water.

When $30ml$ of $H_2$ will react with $15ml$ of $O_2$ to form water it can be represented in a balanced chemical reaction as:

$\underset{0}{\underset{30}{H_2\left(g\right)}}+\underset{(20-15)}{\underset{20}{\frac{1}{2}{O}_2\left(g\right)}}\to \underset{30}{\underset{0}{H_{2}O\left(g\right)}}$

So at the end of the reaction,$30$ ml of water and $5ml$ of $O_2$ and $10ml$ of $H_2$ will be left.

Hence, the correct option is $A$ and $D$