\(\begin{array}{l}
Bao\,toan\,khoi\,luong:\\
m_A+m_{O_2}=m_{oxit}\\
\to m_{O_2}=28,4-15,6=12,8(g)\\
\to n_{O_2}=\frac{12,8}{32}=0,4(mol)\\
Bao\,toan\,O:\,n_{H_2O}=2n_{O_2}=0,8(mol)\\
Bao\,toan\,H:\,n_{H_2}=n_{H_2O}=0,8(mol)\\
\to V_{H_2}=0,8.22,4=17,92(l)
\end{array}\)