$a/Cu+2H_2SO_4\overset{t^o}\to 2H_2O+SO_2+CuSO_4$
$2Al+6H_2SO_4\overset{t^o}\to Al_2(SO_4)_3+6H_2O+3SO_2$
$b/n_{SO_2}=2,24/22,1=0,1mol$
Gọi $n_{Cu}=a;n_{Al}=b$
Ta có :
$m_{hh}=64a+27b=5,5$
$n_{SO_2}=a+1,5b=0,1$
Ta có hpt :
$\left\{\begin{matrix}
64a+27b=5,5 & \\
a+1,5b=0,1 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a≈0,08 & \\
b≈0,01 &
\end{matrix}\right.$
Theo pt :
$n_{CuSO_4}=n_{Cu}=0,08mol$
$n_{Al_2(SO_4)_3}=1/2.n_{Al}=1/2.0,01=0,005mol$
$⇒m=0,08.160+0,005.342=14,51g$