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

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

Theo phương trình ta có:

${{n}_{{{C}_{6}}{{H}_{6}}}}={{n}_{B{{r}_{2}}}}=0,15\left( mol \right)$

${{V}_{{{C}_{6}}{{H}_{6}}}}=\frac{{{n}_{{{C}_{6}}{{H}_{6}}}}}{{{C}_{{{M}_{{{C}_{6}}{{H}_{6}}}}}}}=\frac{0,15}{1}=0,15\left( l \right)=150ml$

$\Rightarrow $Chọn đáp án C