$4H^+ + SO_4^{2-}+2e^- \to SO_2+2H_2O $

`=>`$n_{SO_4}=n_{SO_2}=\dfrac{9.072}{22.4}=0.405(mol)$

$m_{muối}=m_{kim \,\, loại}+m_{SO_4}$

`<=>`$53.79=m+96 \times 0.405$

`=>`$m=14.91(gam)$

Gọi $n_{Mg}=a(mol);n_{Cu}=b(mol);n_{Al}=c(mol)$

`->`$24a+64b+27c=14.91$ $(1)$

Mà $\%n_{Al}=25\%$

`=>`$\dfrac{c}{a+b+c}=0.25$

`<=>`$a+b-3c=0$   $(2)$

Bảo toàn $e:$

$2n_{Mg}+2n_{Cu}+3n_{Al}=2n_{SO_2}$

`<=>`$2a+2b+3c=0.81$ $(3)$

Từ $(1),(2)$ và $(3)$ suy ra: $a=0.12;b=0.15;c=0.09$

`=>`$m_{Al}=0.09 \times 27=2.43(gam)$