Hướng dẫn:

${{n}_{FeC{{l}_{2}}}}=\frac{{{m}_{FeC{{l}_{2}}}}}{{{M}_{FeC{{l}_{2}}}}}=\frac{12,7}{127}=0,1\left( mol \right)$ 

$FeC{{l}_{2}}+2AgN{{O}_{3}}\to Fe{{\left( N{{O}_{3}} \right)}_{2}}+2AgC{{l}_{\downarrow }}$ 

 0,1                       $\to $                       0,2           mol

Theo phương trình ta có: ${{n}_{AgCl}}=2.{{n}_{FeC{{l}_{2}}}}=2.0,1=0,2\left( mol \right)$

${{m}_{AgCl}}={{n}_{AgCl}}.{{M}_{AgCl}}=0,2.143,5=28,7\left( g \right)$ 

Chọn đáp án B