`1)` Cho $\begin{cases} Mg: x(mol)\\ Al: y(mol)\\\end{cases}$
`=> 24x+27y=7,8g(1)`
`n_{H_2}=\frac{8,96}{22,4}=0,4(mol)`
BTe:
$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e\ || \ \mathop{2H}\limits^{+}+2e\to \mathop{H_2}\limits^{0}$
$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$
`=> 2x+3y=0,8(mol)(2)`
`(1),(2)=> x=0,1(mol), y=0,2(mol)`
`=> m_{Mg}=0,1.24=2,4g`
`m_{Al}=27.0,2=5,4g`
`2)` `n_{H_2}=\frac{4,48}{22,4}=0,2(mol)`
$\begin{cases} Mg: x(mol)\\ Fe: y(mol)\\\end{cases}$
`=> 24x+56y=8g(1)`
BTe:
$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e\ || \ \mathop{2H}\limits^{+}+2e\to \mathop{H_2}\limits^{0}$
$\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+2}+2e$
`2x+2y=0,4(mol)(2)`
`(1),(2)=>x=y=0,1(mol)`
`=> %m_{Mg}=\frac{2,4.100%}{8}=30%`
`%m_{Fe}=100%-30%=70%`
`3)` `n_{SO_2}=\frac{1,12}{22,4}=0,05(mol)`
`Cu+2H_2SO_{4 \text{đ, nguội}}\to CuSO_4+2H_2O+SO_2`
`=> n_{SO_2}=n_{Cu}=0,05(mol)`
`=> m_{Cu}=0,05.64=3,2g`
`=> m_{Al}=14,4-3,2=11,2g`
