Hướng dẫn:

${{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\frac{{{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}=\frac{180.100}{180}=100\left( mol \right)$

${{n}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{{{m}_{{{C}_{2}}{{H}_{5}}OH}}}{{{M}_{{{C}_{2}}{{H}_{5}}OH}}}=\frac{8,28.1000}{46}=180\left( mol \right)$

                     ${{C}_{6}}{{H}_{12}}{{O}_{6}}\xrightarrow{30-32{}^\circ C}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}}\uparrow $

Ban đầu:           100                                                                 (mol)

Phản ứng:         90             $\leftarrow $          180                                  (mol)

Theo phương trình ta có:

${{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}pu}}=\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{2}=\frac{180}{2}=90\left( mol \right)$

Hiệu suất của quá trình:

 $%H=\frac{{{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6pu}}}}}{{{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}.100=\frac{90}{100}.100=90\left( % \right)$

$\Rightarrow $Chọn đáp án C