Em tham khảo nha :
\(\begin{array}{l}
2{H_2} + {O_2} \to 2{H_2}O\\
2CO + {O_2} \to 2C{O_2}\\
{n_{{O_2}}} = \dfrac{{89,6}}{{22,4}} = 4mol\\
hh:{H_2}(a\,mol);Co(b\,mol)\\
\left\{ \begin{array}{l}
2a + 28b = 68\\
\dfrac{a}{2} + \dfrac{b}{2} = 4
\end{array} \right.\\
\Rightarrow a = 6;b = 2\\
{m_{{H_2}}} = 6 \times 2 = 12g\\
{m_{CO}} = 68 - 12 = 56g\\
\% {V_{{H_2}}} = \dfrac{6}{{6 + 2}} \times 100\% = 75\% \\
\% {V_{CO}} = 100 - 75 = 25\%
\end{array}\)