\(\begin{array}{l}
a)\\
Dat\,n_{Mg}=x(mol);n_{Fe}=y(mol);n_{Zn}=z(mol)\\
\to 24x+56y+65z=22,1(1)\\
Mg+H_2SO_4\to MgSO_4+H_2\\
Fe+H_2SO_4\to FeSO_4+H_2\\
Zn+H_2SO_4\to ZnSO_4+H_2\\
Theo\,PT:\,x+y+z=n_{H_2}=\frac{12,32}{22,4}=0,55(2)\\
Ta\,co\,n_{H_2(Mg)}=2n_{H_2(Fe)}\\
\to x=2y(3)\\
(1)(2)(3)\to x=0,3;y=0,15;z=0,1\\
\to m_{Mg}=0,3.24=7,2(g);m_{Fe}=0,15.56=8,4(g)\\
m_{Zn}=0,1.65=6,5(g)\\
b)\\
Theo\,PT:\,n_{MgSO_4}=0,3(mol);n_{FeSO_4}=0,15(mol);n_{ZnSO_4}=0,1(mol)\\
\to m_B=0,3.120+0,15.152+0,1.161=74,9(g)
\end{array}\)

`2,`

PTHH:

`Mg+2HCl->MgCl_2+H_2(1)`

`Fe+2HCl->FeCl_2+H_2(2)`

`Zn+2HCl->ZnCl_2+H_2(3)`

`n_{H_2}=(12,32)/(22,4)=0,55(mol)`

Gọi số mol `H_2` tạo thành do `Fe,Zn` lần lượt là `a,b`

`=>` Số mol `H_2` tạo thành do `Mg` là : `2a`

Ta có : `a+b+2a=0,55(I)`

Ta lại có theo PTHH ta suy ra :

`n_{Mg}=2a(mol);n_{Fe}=a(mol);n_{Zn}=b(mol)`

`=>24.2a+56a+65b=104a+65b=22,1(II)`

Từ (I) và (II) ta suy ra :

`a=0,15;b=0,1`

`a,`

`n_{Mg}=2a=2.0,15=0,3(mol)`

`n_{Fe}=a=0,15(mol)`

`n_{Zn}=b=0,1(mol)`

`m_{Mg}=0,3.24=7,2(g)`

`m_{Fe}=0,15.56=8,4(g)`

`m_{Zn}=0,1.65=6,5(g)`

`b,`

Theo PTHH :

`n_{MgSO_4}=n_{Mg}=0,3(mol)`

`n_{FeSO_4}=n_{Fe}=0,15(mol)`

`n_{ZnSO_4}=n_{Zn}=0,1(mol)`

`=>m_{m uối}=0,3.120+0,15.152+0,1.161=74,9(g)`