Hướng dẫn:

$AgN{{O}_{3}}+NaCl\to AgCl+NaN{{O}_{3}}$

$\begin{align}  & {{n}_{AgN{{O}_{3}}}}=\frac{{{m}_{AgN{{O}_{3}}}}}{{{M}_{AgN{{O}_{3}}}}}=\frac{8,5}{170}=0,05mol \\ & {{n}_{NaCl}}=\frac{{{m}_{NaCl}}}{{{M}_{NaCl}}}=\frac{1,755}{58,5}=0,03mol \\\end{align}$

Theo phương trình ta có:${{n}_{AgN{{O}_{3}}}}={{n}_{NaCl}}$

Suy ra: NaCl phản ứng hết.

$\begin{align}  & {{n}_{AgCl}}={{n}_{NaCl}}=0,03mol \\ & {{m}_{AgCl}}={{n}_{AgCl}}.{{M}_{AgCl}}=0,03.143,5=4,305g \\\end{align}$

                     $\Rightarrow $ Chọn đáp án B.