${{n}_{B{{r}_{2}}}}={{V}_{B{{r}_{2}}}}.{{C}_{{{M}_{B{{r}_{2}}}}}}=0,1.2=0,2\left( mol \right)$

${{C}_{2}}{{H}_{2}}+2B{{r}_{2}}\to {{C}_{2}}{{H}_{2}}B{{r}_{4}}$

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

Theo phương trình ta có:

${{n}_{{{C}_{2}}{{H}_{2}}}}=\frac{{{n}_{B{{r}_{2}}}}}{2}=\frac{0,2}{2}=0,1\left( mol \right)$

$\Rightarrow {{V}_{{{C}_{2}}{{H}_{2}}}}={{n}_{{{C}_{2}}{{H}_{2}}}}.22,4=0,1.22,4=2,24\left( l \right)$

$\Rightarrow $Chọn đáp án A