Hướng dẫn :

${{n}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=\frac{{{m}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}}{{{M}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}}=\frac{17,1}{342}=0,05mol$

${{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\frac{{{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}=\frac{7,56}{180}=0,042mol$

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

Theo phương trình phản ứng ta có: ${{n}_{Glucozo}}={{n}_{Saccarozo}}$

$\Rightarrow $ Hiệu suất phản ứng thủy phân là:

$H=\frac{{{n}_{saccaroz{{o}_{pu}}}}}{{{n}_{saccaroz{{o}_{bd}}}}}.100$%$=\frac{0,042}{0,05}.100=84$%

                      $\Rightarrow $ Chọn đáp án A.