\(n_{CuSO_4}=0,2.1=0,2(mol)\\ PTHH:Mg+CuSO_4\to MgSO_4+Cu\\ \Rightarrow n_{Mg}=0,2(mol)\\ \Rightarrow m_{Mg}=0,2.24=4,8(g)\\ \Rightarrow m_{Ag}=m_{hh}-m_{Mg}=20-4,8=15,2(g)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

a) Theo Pt : \(n_{H2}=n_{Fe}=n_{FeCl2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\)

\(\%m_{Cu}=100\%-56\%=44\%\)

b) Theo Pt : \(n_{H2}=2n_{HCl}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{7,3\%}.100\%=100\left(g\right)\)

c) \(m_{ddspu}=10+100-0,1.2=109,8\left(g\right)\)

\(C\%_{FeCl2}=\dfrac{0,1.127}{109,8}.100\%=11,57\%\)

\(n_{Zn}=n_{FeSO_4}=0.2\left(mol\right)\)

\(\Rightarrow m_{Zn}=0.2\cdot65=13\left(g\right)\)

\(m_{Cu}=20-13=7\left(g\right)\)

\(B\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow m_{Cu}=12-5,6=6,4\left(g\right)\)

Nhận thấy rừng `Cu` không tác dụng với `HCl` nên toàn bộ lượng `H_2` là do `Fe`

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

tỉ lệ         1    :       2      :         1     :      1

n(mol)    0,1<------------------------------0,1

\(m_{Fe}=n\cdot M=0,1\cdot56=5,6\left(g\right)\\ m_{Cu}=12-5,6=6,4\left(g\right)\)

\(n_{Mg}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(m_{hh}=24a+27b=7.5\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+H_2\)

\(n_{H_2}=a+1.5b=0.35\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.1\)

\(\%Mg=\dfrac{0.2\cdot24}{7.5}\cdot100\%=64\%\)

\(\%Al=100\%-64\%=36\%\)