$n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{5,6}{22,4}=0,25(mol)$
$n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{3,36}{22,4}=0,15(mol)$
$2H_2+O_2→2H_2O$
$2$ $:$ $1$ $:$ $2$
$\dfrac{0,25}{2}<\dfrac{3}{20}$
$⇒O_2$ dư
$⇒n_{H_2O}=0,25.2:2=0,25(mol)$
$⇒m_{H_2O}=n_{H_2O}.M{H_2O}=0,25.(1.2+16)=4,5(g)$