Hướng dẫn:

$2Al+3C{{l}_{2}}\to 2AlC{{l}_{3}}(1)$

$AlC{{l}_{3}}+3AgN{{O}_{3}}\to Al{{(N{{O}_{3}})}_{3}}+3AgCl(2)$ 

${{n}_{AgCl}}=\frac{{{m}_{AgCl}}}{{{M}_{AgCl}}}=\frac{2,583}{143,5}=0,018$ (mol)

Theo phương trình (2): ${{n}_{AlC{{l}_{3}}}}=\frac{1}{3}{{n}_{AgCl}}=\frac{1}{3}.0,018=0,006$ (mol)

Theo phương trình (1): ${{n}_{Al}}={{n}_{AlC{{l}_{3}}}}=0,006$(mol)

${{m}_{Al}}={{n}_{Al}}.{{M}_{Al}}=0,006.27=0,162$ g

                      \[\Rightarrow \] Chọn đáp án B.