\(n_{Zn} = a(mol) ; n_{Al} = b(mol) ; n_{Mg} = c(mol)\\ \Rightarrow 65a + 27b + 24c = 44,1(1)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3 H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{31,36}{22,4} = 1,4(2)\\ Mà : 2a = 3b(3)\\ (1)(2)(3) \Rightarrow a = 0,3 ; b = 0,2 ; c = 0,8\\ \%m_{Zn} = \dfrac{0,3.65}{44,1}.100\% = 44,22\%\\ \%m_{Al} = \dfrac{0,2.27}{44,1}.100\% = 12,24\%\)

\(\%m_{Mg} = 100\% -44,22\% -12,24\% = 43,54\%\)

\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}65x+24y=12,5\\x+y=0,35\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,25\end{matrix}\right.\\ \Rightarrow m_{Zn}=6,5\left(g\right);m_{Mg}=6\left(g\right)\\ b.Tacó:BTNT\left(H\right):n_{HCl}.1>n_{H_2}.2\\ \Rightarrow HCldưsauphảnứng\\ Dungdịchsauphảnứnggồm:\left\{{}\begin{matrix}ZnCl_2:0,1\left(mol\right)\\MgCl_2:0,25\left(mol\right)\\HCl_{dư}:0,8-0,7=0,1\left(mol\right)\end{matrix}\right.\\ m_{ddsaupu}=200+12,5-0,35.2=212,8\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,1.136}{212,8}.100=6,39\%;C\%_{MgCl_2}=\dfrac{0,25.95}{212,8}.100=11,16\%;C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{212,8}.100=1,72\%\)

a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+27b=5,1\)  (1)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Bảo toàn electron: \(2a+3b=0,5\)  (2)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%\approx47,06\%\\\%m_{Al}=52,94\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{7,3\%}=250\left(g\right)\)

\(\Rightarrow V_{HCl}=\dfrac{250}{1,2}\approx208,33\left(ml\right)\)

\(n_{H_2}=\dfrac{4,368}{22,4}=0,195mol\)

\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+24y=3,87\left(1\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

\(\Rightarrow1,5x+y=0,195\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,09\\y=0,06\end{matrix}\right.\)

\(m_{Al}=0,09\cdot27=2,43g\)

\(m_{Mg}=0,06\cdot24=1,44g\)

\(a) n_{Mg}= a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b =2,55(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{2,8}{22,4}=0,125(2)\\ (1)(2) \Rightarrow a = b = 0,05\\ \%m_{Mg} = \dfrac{0,05.24}{2,55}.100\% = 47,06\%\ ;\ \%m_{Al} =100\% -47,06\% = 52,94\%\\ b) n_{HCl} = 2n_{H_2} = 0,125.2 = 0,25(mol)\\ m_{dd\ HCl} = \dfrac{0,25.36,5}{7,3\%} = 125(gam)\\ V_{dd\ HCl} = \dfrac{125}{1,2} = 104,17(ml)\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5(mol)\\ n_{Mg}=x(mol);n_{Fe}=y(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow \begin{cases} 24x+56y=20\\ x+y=0,5\\ \end{cases} \Rightarrow x=y=0,25(mol)\\ \Rightarrow \begin{cases} m_{Fe}=56.0,25=14(g)\\ m_{Mg}=24.0,25=6(g)\\ \end{cases}\)

a/ Gọi số mol Al, Mg trong hỗn hợp là a, b

PTHH:

2Al + 6HCl ===> 2AlCl₃ + 3H₂

a............................................1,5a

Mg + 2HCl ===> MgCl₂ + H₂

b.........................................b

n_H2 = 5,6 / 22,4 = 0,25 (mol)

The đề ra, ta có hệ phương trình:

\(\begin{cases}27a+24b=5,1\\1,5a+b=0,25\end{cases}\)=> \(\begin{cases}a=0,1\\b=0,1\end{cases}\)

=> m_Al = 0,1 x 27 = 2,7 gam

m_Mg = 0,1 x 24 = 2,4 gam

=> %m_Al = \(\frac{2,7}{5,1}.100\%=52,94\%\)

%m_Mg = 100% - 52,94% = 46,06%

b/ Tổng số mol của HCl = 0,3 + 0,2 = 0,5 mol

=> m_HCl = 0,5 x 36,5 = 18,25 gam

c/ Áp dụng định luật bảo toàn khối lượng, ta có
m_{hỗn hợp muối} = m_{kim loại} + m_HCl - m_H2

₌ 5,1 + 18,25 - 0,25 x 2 = 22,85 gam