${{n}_{{{C}_{6}}{{H}_{5}}Br}}=\frac{{{m}_{{{C}_{6}}{{H}_{5}}Br}}}{{{M}_{{{C}_{6}}{{H}_{5}}Br}}}=\frac{23,55}{157}=0,15\left( mol \right)$

$$

   ${{C}_{6}}{{H}_{6}}+B{{r}_{2}}\underrightarrow{Fe,{{t}^{0}}} &  &  & {{C}_{6}}{{H}_{5}}Br+HBr$

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

Theo phương trình ta có:

${{n}_{{{C}_{6}}{{H}_{6}}}}={{n}_{{{C}_{6}}{{H}_{5}}Br}}=0,15\left( mol \right)$

$\Rightarrow {{m}_{{{C}_{6}}{{H}_{6}}}}={{n}_{{{C}_{6}}{{H}_{6}}}}.{{M}_{{{C}_{6}}{{H}_{6}}}}=0,15.78=11,7\left( g \right)$

$\Rightarrow $Chọn đáp án C