Hướng dẫn:

${{n}_{C}}=\frac{{{m}_{C}}}{{{M}_{C}}}=\frac{2,4}{12}=0,2\left( mol \right)$

$C{{O}_{2}}+2Mg\xrightarrow{{{t}^{0}}}2MgO+C$
0,2               $\leftarrow $                  0,2      (mol)
Theo phương trình ta có: ${{n}_{C{{O}_{2}}}}={{n}_{C}}=0,2\left( mol \right)$

$\Rightarrow {{V}_{C{{O}_{2}}}}=22,4.{{n}_{C{{O}_{2}}}}=22,4.0,2=4,48\left( l \right)$

$\Rightarrow $ Chọn đáp án D