${{m}_{{{C}_{6}}{{H}_{6}}}}={{m}_{A}}.75%=26.75%=19,5\left( g \right)$

$\Rightarrow {{n}_{{{C}_{6}}{{H}_{6}}}}=\frac{{{m}_{{{C}_{6}}{{H}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{6}}}}}=\frac{19,5}{78}=0,25\left( mol \right)$

   ${{C}_{6}}{{H}_{6}}+3{{H}_{2}}\underrightarrow{Ni,{{t}^{0}}} & {{C}_{6}}{{H}_{12}}$

  0,25 $\to $0,75                             (mol)

Theo phương trình ta có:

${{n}_{{{H}_{2}}}}=3.{{n}_{{{C}_{6}}{{H}_{6}}}}=3.0,25=0,75\left( mol \right)$

$\Rightarrow {{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,75=16,8\left( l \right)$

$\Rightarrow $Chọn đáp án D