$a)$ Đặt $(n_{Fe};n_{Al})=(x;y)(mol)$

$\Rightarrow 56x+27y=11(1)$

$PTHH:Fe+H_2SO_4\to FeSO_4+H_2\uparrow$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\uparrow$

Theo 2 PT: $x+1,5y=n_{H_2}=\dfrac{8,96}{22,4}=0,4(2)$

Từ $(1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)$

$\Rightarrow \%m_{Fe}=\dfrac{0,1.56}{11}.100\%=50,91\%$

$\Rightarrow \%m_{Al}=100\%-50,91\%=49,09\%$

$b)$ Theo 2 PT: $n_{FeSO_4}=x=0,1(mol);n_{Al_2(SO_4)_3}=0,5.y=0,1(mol)$

$\Rightarrow m_{FeSO_4}=0,1.152=15,2(g)$

$\Rightarrow m_{Al_2(SO_4)_3}=0,1.342=34,2(g)$

$c)$ Đặt CTHH oxit KL là $M_aO_b$

$PTHH:M_aO_b+bH_2\xrightarrow{t^o}aM+bH_2O$

Theo PT: $n_{M_aO_b}=\dfrac{n_{H_2}}{b}=\dfrac{0,4}{b}$

$\Rightarrow M_{M_aO_b}=\dfrac{23,2}{\dfrac{0,4}{b}}=58b(g/mol)$

$\Rightarrow M_M.x+16y=58y$

$\Rightarrow M_M.x=42y$

$\Rightarrow \dfrac{x}{y}=\dfrac{42}{M_M}$

Với $x=3;y=4\Rightarrow M_M=56(g/mol)(Fe)$

Vậy CTHH oxit là $Fe_3O_4$