\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:H_2+2O_2\xrightarrow{t^o}2H_2O\\ \Rightarrow n_{H_2O}=2n_{H_2}=0,2(mol)\\ \Rightarrow m_{H_2O}=0,2.18=3,6(g)\)
Bảo toàn nguyên tố H:
\(n_{H_2O}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2O}=1,8\left(g\right)\)