Hướng dẫn:

${{n}_{Ag}}=\frac{{{m}_{Ag}}}{{{M}_{Ag}}}=\frac{12,96}{108}=0,12\left( mol \right)$

${{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag\downarrow $

  0,06                       $\leftarrow $                      0,12                 (mol)

Theo phương trình ta có:

${{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\frac{{{n}_{Ag}}}{2}=\frac{0,12}{2}=0,06\left( mol \right)$

${{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}={{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}.{{M}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=0,06.180=10,8\left( g \right)$

$\Rightarrow $Chọn đáp án A