Đáp án đúng: C
33,33% và 66,67%
$\left\{ \begin{array}{l}44{{n}_{C{{O}_{2}}}}\,+\,28{{n}_{CO}}\,=\,{{m}_{C}}\,+\,{{m}_{{{O}_{2}}}}=3,6\,+\,\frac{4,48}{22,4}.32\\{{n}_{C{{O}_{2}}}}\,+\,{{n}_{CO}}\,=\,{{n}_{C}}\,=\frac{3,6}{12}\,=0,3\end{array} \right.$
$\Rightarrow {{n}_{C{{O}_{2}}}}\,=\,0,1\,mol;\,\,{{n}_{CO}}\,=\,0,2\,mol$
$\Rightarrow {\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}{{V}_{C{{O}_{2}}}}\,=\,\frac{0,1}{0,3}\,=\,33,33{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}$