Hướng dẫn:

${{n}_{Al}}=\frac{{{m}_{Al}}}{{{M}_{Al}}}=\frac{1,08}{27}=0,04$ (mol)

${{n}_{Cu}}=\frac{{{m}_{Cu}}}{{{M}_{Cu}}}=\frac{2,88}{64}=0,045$ (mol)

$2Al+3CuC{{l}_{2}}\to 2AlC{{l}_{3}}+3Cu$ 

Theo phương trình: ${{n}_{A{{l}_{pu}}}}=\frac{2}{3}{{n}_{Cu}}=\frac{2}{3}.0,045=0,03$(mol)

${{n}_{A{{l}_{du}}}}={{n}_{Al}}-{{n}_{A{{l}_{pu}}}}=0,04-0,03=0,01$ (mol)

${{m}_{A{{l}_{du}}}}=0,01.27=0,27$ (g)

                      \[\Rightarrow \] Chọn đáp án A.