\(n_{H_2}:\dfrac{17,92}{22,4}=0,8\left(mol\right)\)
\(n_{O_2}:\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(2H_2+O_2\underrightarrow{t^o}2H_2O\)
2............1.........2(mol)
0,8.......0,4........0,8(mol)
=> Oxi dư
\(m_{H_2O}:0,8.18=14,4\left(g\right)\)
\(V_{H_2O}:\dfrac{14,4}{1}=14,4\left(ml\right)\)