${{n}_{C{{O}_{2}}}}=\frac{{{V}_{C{{O}_{2}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right)$

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

0,075                 $\leftarrow $    0,15                         (mol)

Theo phương trình ta có: ${{n}_{{{C}_{2}}{{H}_{2}}}}=\frac{2.{{n}_{C{{O}_{2}}}}}{4}=\frac{2.0,15}{4}=0,075\left( mol \right)$

$\Rightarrow {{V}_{{{C}_{2}}{{H}_{2}}}}={{n}_{{{C}_{2}}{{H}_{2}}}}.22,4=0,075.22,4=1,68\left( l \right)$

$\Rightarrow $Chọn đáp án A