${{m}_{B{{r}_{2}}}}=\frac{{{m}_{\text{d}{{\text{d}}_{B{{r}_{2}}}}}}.C%}{100}=\frac{500.8}{100}=40\left( g \right)$

  $\Rightarrow {{n}_{B{{r}_{2}}}}=\frac{{{m}_{B{{r}_{2}}}}}{{{M}_{B{{r}_{2}}}}}=\frac{40}{160}=0,25\left( mol \right)$

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

  0,125 $\leftarrow $   0,25                              (mol)

Theo phương trình ta có:

${{n}_{{{C}_{2}}{{H}_{2}}}}=\frac{{{n}_{B{{r}_{2}}}}}{2}=\frac{0,25}{2}=0,125\left( mol \right)$

${{V}_{{{C}_{2}}{{H}_{2}}}}=22,4.{{n}_{{{C}_{2}}{{H}_{2}}}}=22,4.0,125=2,8\left( l \right)$

$\Rightarrow $Chọn đáp án D