Hướng dẫn :

${{n}_{Ag}}=\frac{27}{108}=0,25$ (mol)

${{C}_{12}}{{H}_{22}}{{O}_{11}}+{{H}_{2}}O\xrightarrow[{{t}^{o}}]{{{H}^{+}}}{{C}_{6}}{{H}_{12}}{{O}_{6}}+{{C}_{6}}{{H}_{12}}{{O}_{6}}(1)$

${{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\xrightarrow{N{{H}_{3}},{{t}^{o}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag(2)$

0,125    $\leftarrow $                                             0,25(mol)

Ta có trong môi trường kiềm fructozo chuyển hóa thành glucozo

Nên ${{n}_{glucoz{{o}_{(pt1)}}}}={{n}_{fructoz{{o}_{(pt1)}}}}=\frac{1}{2}{{n}_{glucoz{{o}_{(pt2)}}}}=\frac{1}{2}.0,125=0,0625$ (mol)

Theo phương trình (1) có ${{n}_{saccarozo}}={{n}_{glucoz{{o}_{(pt1)}}}}=0,0625$ (mol)

$\to a=0,0625.342=21,375g$

                      $\Rightarrow $ Chọn đáp án B.