\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\)
\(0.05...0.025......0.05\)
\(m_{H_2O}=0.05\cdot18=0.9\left(g\right)\)
\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{H_2O} = n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)\\ m_{H_2O} = 0,05.18 = 0,9(gam)\)