\(2Al+6HCl\to 2AlCl_3+3H_2\\ Fe+2HCl\to FeCl_2+H_2\\ n_{HCl}=0,2.4=0,8(mol)\\ \Rightarrow \begin{cases} 56.n_{Fe}+27.n_{Al}=22\\ 2.n_{Fe}+3.n_{Al}=0,8 \end{cases}\Rightarrow \begin{cases} n_{Fe}=0,39(mol)\\ n_{Al}=0,007(mol) \end{cases}\\ \Rightarrow \begin{cases} \%m_{Fe}=\dfrac{0,39.56}{22}.100\%=99,27\%\\ \%m_{Al}=100\%-99,27\%=0,73\% \end{cases}\)

Đặt n_CuO = x (mol) ; n_Fe2O3 = y (mol)

CuO + HCl -------> CuCl₂ + H₂O

x-------> x (mol)

Fe₂O₃ + 6HCl -------------> 2FeCl3 + 3H₂O

y--------->6y (mol)

Theo đề bài ta có hệ

\(\left\{{}\begin{matrix}80x+160y=20\\x+6y=0,2.3,5\end{matrix}\right.\)

=> x = 0,025 (mol) ; y = 0,1125 (mol)

%m_CuO = 0,025 . 80 .100% / 20 = 10 %

%m_Fe2O3 = 100 % - 10 % = 90 %

b) 20 % và 80%

Gọi \(n_{CuO} và n_{Fe_2O_3} là x y\)

\(\begin{cases} 80x + 160y= 20\\ 2x+ 6y= 0,2 . 3,5=0,7 \end{cases} \)

x=0,05

y= 0,1

%\(m_{CuO}= \dfrac{0,05 . 80}{20} . 100\)%= 20%

=> %m_{\(Fe_2O_3\)}= 80%

Giả sử trong mỗi phần có: \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\\n_{Zn}=c\left(mol\right)\\n_O=d\left(mol\right)\end{matrix}\right.\)

=> 56a + 64b + 65c + 16d = 32,21 

P1: 

n_O = n_H2O = d (mol)

=> n_HCl = 2d (mol)

Theo ĐLBTKL: m_{rắn bđ} + m_HCl = m_muối + m_H2O

=> 32,21 + 73d = 59,16 + 18d

=> d = 0,49 (mol)

P2: 

Gọi số mol HCl, H₂SO₄ là a, b (mol)

n_H2O = n_O = 0,49 (mol)

Bảo toàn H: a + 2b = 0,98  (1)

Theo ĐLBTKL: m_{rắn bđ} + m_HCl + m_H2SO4 = m_muối + m_H2O

=> 32,21 + 36,5a + 98b = 65,41 + 0,49.18

=> 36,5a + 98b = 42,02 (2)

(1)(2) => a = 0,48 (mol); b = 0,25 (mol)

=> \(\left\{{}\begin{matrix}C_{M\left(HCl\right)}=\dfrac{0,48}{1}=0,48M\\C_{M\left(H_2SO_4\right)}=\dfrac{0,25}{1}=0,25M\end{matrix}\right.\)

\(n_{HCl}=0,2.3,5=0,7\left(mol\right)\\ n_{CuO}=a;n_{Fe_2O_3}=b\\ CuO+2HCl\xrightarrow[]{}\Rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\xrightarrow[]{}2FeCl_3+H_2O\\ \left\{{}\begin{matrix}2a+6b=0,7\\80a+160b=20\end{matrix}\right.\\\Rightarrow a=0,05;b=0,1\\ \%_{CuO}=\dfrac{0,05.80}{20}\cdot100=20\%\\ \%_{Fe_2O_3}=\dfrac{0,1.160}{20}\cdot100=80\%\)

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\)

a)\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

     0,1             0,2        0,1         0,1         0,1

b)\(m_{HCl}=0,1\cdot36,5=3,65\left(g\right)\)

   \(a\%=\dfrac{3,65}{100}\cdot100\%=3,65\%\)

c)\(m_{CaCO_3}=0,1\cdot100=10\left(G\right)\)

  \(\Rightarrow\%m_{CaCO_3}=\dfrac{10}{16}\cdot100\%=62,5\%\)

  \(\Rightarrow\%m_{CaCl_2}=100\%-62,5\%=37,5\%\)

d)\(m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\)

   \(m_{H_2O}=0,1\cdot18=1,8\left(g\right)\)

   \(m_{ddsau}=10+100-0,1\cdot44-1,8=103,8\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{11,1}{103,8}\cdot100\%=10,7\%\)

a) Gọi số mol Mg, CuO là a, b (mol)

=> 24a + 80b = 14 (1)

\(n_{HCl}=\dfrac{255,5.10\%}{36,5}=0,7\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             a--->2a--------->a------>a

           CuO + 2HCl --> CuCl₂ + H₂O

               b------>2b----->b

=> 2a + 2b = 0,7 (2)

(1)(2) => a = 0,25 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{14}.100\%=42,857\%\\\%m_{CuO}=\dfrac{0,1.80}{14}.100\%=57,143\%\end{matrix}\right.\)

b)

m_{dd sau pư} = 14 + 255,5 - 0,25.2 = 269 (g)

\(C\%_{MgCl_2}=\dfrac{0,25.95}{269}.100\%=8,829\%\)

\(C\%_{CuCl_2}=\dfrac{0,1.135}{269}.100\%=5,019\%\)