Hướng dẫn :

${{n}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=\frac{136,8}{342}=0,4mol$

${{C}_{12}}{{H}_{22}}{{O}_{11}}+{{H}_{2}}O\xrightarrow[{{t}^{o}}]{{{H}^{+}}}{{C}_{6}}{{H}_{12}}{{O}_{6}}+{{C}_{6}}{{H}_{12}}{{O}_{6}}$

0,4                     $\to $ $0,4.0,9=0,36$ (mol) ($H=90$% )

${{C}_{6}}{{H}_{12}}{{O}_{6}}\xrightarrow[ruou,{{t}^{o}}]{men}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}}$

0,36        $\to $  $2.0,36.0,9=0,648$ (mol)      ($H=90$% )

$\Rightarrow {{m}_{{{C}_{2}}{{H}_{5}}OH}}=0,648.46=29,808g$

                      $\Rightarrow $ Chọn đáp án C.