Hướng dẫn:

Gọi ${{n}_{CuS{{O}_{4}}}}=a$ ;${{n}_{AgN{{O}_{3}}}}=b$ .

 $2Al+3CuS{{O}_{4}}\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+3Cu(1)$

$\frac{2}{3}a$       $\leftarrow $  a  $\to $                    a

$Al+3AgN{{O}_{3}}\to Al{{(N{{O}_{3}})}_{3}}+3Ag(2)$

$\frac{1}{3}b$    $\leftarrow $      b $\to $               b

theo bài cho :$\begin{align}& a+b=0,09 \\ & 64a+108b=7,96 \\ \end{align}$

                    $\Rightarrow a=0,04;b=0,05$ (mol)

Theo phương trình: ${{n}_{Al}}=\frac{2}{3}a+\frac{1}{3}b=\frac{2}{3}.0,04+\frac{1}{3}.0,05=\frac{13}{300}$ (mol)

${{m}_{Al}}={{n}_{Al}}.{{M}_{Al}}=\frac{13}{300}.27=1,17g$ 

                      Chọn đáp án C.