Hướng dẫn:

${{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\frac{{{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}=\frac{13,5}{180}=0,075\left( mol \right)$

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

  0,075                  $\to $                         0,15                 (mol)

Theo phương trình ta có:

${{n}_{Ag}}=2.{{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=2.0,075=0,15\left( mol \right)$

$\Rightarrow {{m}_{Ag}}={{n}_{Ag}}.{{M}_{Ag}}=0,15.108=16,2\left( g \right)$

$\Rightarrow $Chọn đáp án A