${{n}_{{{C}_{2}}{{H}_{2}}}}=\frac{{{V}_{{{C}_{2}}{{H}_{2}}}}}{22,4}=\frac{2,24}{22,4}=0,1\left( mol \right)$

$2{{C}_{2}}{{H}_{2}}+5{{O}_{2}}\xrightarrow{{{t}^{0}}}4C{{O}_{2}}+2{{H}_{2}}O$

  0,1               $\to $     0,2                        (mol)

Theo phương trình ta có:

${{n}_{C{{O}_{2}}}}=\frac{4.{{n}_{{{C}_{2}}{{H}_{2}}}}}{2}=\frac{4.0,1}{2}=0,2\left( mol \right)$

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

$\Rightarrow $Chọn đáp án D