${{n}_{C{{H}_{4}}}}=\frac{{{V}_{C{{H}_{4}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right)$

${{n}_{{{C}_{2}}{{H}_{2}}}}=\frac{{{V}_{{{C}_{2}}{{H}_{2}}}}}{22,4}=\frac{5,6}{22,4}=0,25\left( mol \right)$

$C{{H}_{4}}+2{{O}_{2}}\xrightarrow{{{t}^{0}}}C{{O}_{2}}+2{{H}_{2}}O$                  (1)

0,15               $\to $ 0,15                 (mol)

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

0,25                 $\to $   0,5                         (mol)

Theo hai phương trình (1) và (2) có:

${{n}_{C{{O}_{2}}}}={{n}_{C{{H}_{4}}}}+\frac{4.{{n}_{{{C}_{2}}{{H}_{2}}}}}{2}=0,15+\frac{4.0,25}{2}=0,65\left( mol \right)$

$\Rightarrow {{V}_{C{{O}_{2}}}}=22,4.{{n}_{C{{O}_{2}}}}=22,4.0,65=14,56\left( l \right)$

$\Rightarrow $Chọn đáp án C