Đặt \(n_{Fe}=x(mol);n_{Mg}=y(mol)\Rightarrow 56x+24y=8(1)\)

\(PTHH:Fe+2HCl\to FeCl_2+H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow 127x+95y=22,2(2)\\ (1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{8}.100\%=70\%\)

Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=9,3(1)\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow x+y=0,15(2)\\ (1)(2)\Rightarrow x=0,05(mol);y=0,1(mol)\\ \Rightarrow \%_{Zn}=\dfrac{0,1.65}{9,3}.100\%=69,89\%\)

Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=18,6(1)\)

\(n_{HCl}=3.0,2=0,6(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow 2x+2y=0,6(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Zn}=0,2.65=13(g)\)

Gọi số mol Fe, Zn là a,b

=> 56a + 65b = 12,1 (1)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

______a---->2a

Zn + 2HCl --> ZnCl₂ + H₂

_b---->2b

=> 2a + 2b = 0,4 (2)

(1)(2) => \(\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

=> \(\%Fe=\dfrac{56.0,1}{12,1}.100\%=46,28\%\)

Đặt \(n_{Fe}=x(mol);n_{Zn}=y(mol)\Rightarrow 56x+65y=12,1(1)\)

\(n_{HCl}=2.0,2=0,4(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow 2x+2y=0,4(2)\\ (1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{HCl}=0,1.1=0,1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) => HCl hết, Fe dư

PTHH: Fe + 2HCl --> FeCl₂ + H₂

___________0,1-------------->0,05_____(mol)

=> V_H2 = 0,05.22,4 = 1,12(l)

tham khảo link

https://cunghocvui.com/phuong-trinh-dieu-che-SiO2

\(n_C=\dfrac{24,64}{22,4}+\dfrac{10,6}{106}=1,2\left(mol\right)\)

\(\Rightarrow m_C=14,4\left(g\right)\)