\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\\
pthh:2H_2O\underrightarrow{\text{đ}p}2H_2+O_2\)
0,2 0,2 0,1
=> \(\left\{{}\begin{matrix}V_{H_2}=0,2.22,4=4,48\left(L\right)\\V_{O_2}=0,1.22,4=2,24\left(L\right)\end{matrix}\right.\)