\(Đặt:a=\%V_{H_2}\left(a>0\right)\\\overline{M}_{hhA}=8,5.2=17\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow 2a+32.\left(100\%-a\right)=17\\ \Leftrightarrow a=50\%\\ \Rightarrow\%V_{H_2}=\%V_{O_2}=50\%\\ \%m_{H_2}=\dfrac{2}{2+32}.100\approx5,882\%\Rightarrow\%m_{O_2}\approx100\%-5,882\%\approx94,118\%\)