Đáp án:

 \({V_{{O_2}}} = 3,36l\)

Giải thích các bước giải:

 \(\begin{array}{l}
2CO + {O_2} \to 2C{O_2}(1)\\
2{H_2} + {O_2} \to 2{H_2}O(2)\\
{n_X} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{{O_2}(1)}} = \dfrac{{{n_{CO}}}}{2}(mol)\\
{n_{{O_2}(2)}} = \dfrac{{{n_{{H_2}}}}}{2}(mol)\\
{n_{{O_2}}} = {n_{{O_2}(1)}} + {n_{{O_2}(2)}} = \dfrac{{{n_{CO}}}}{2} + \dfrac{{{n_{{H_2}}}}}{2}\\
 \Rightarrow {n_{{O_2}}} = \dfrac{{{n_{CO}} + {n_{{H_2}}}}}{2} = \dfrac{{0,3}}{2} = 0,15mol\\
{V_{{O_2}}} = n \times 22,4 = 0,15 \times 22,4 = 3,36l
\end{array}\)