\(Đặt:n_{Al}=u\left(mol\right);n_{Fe}=v\left(mol\right)\left(u,v>0\right)\\ n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56u=22\\1,5a+u=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,4\\u=0,2\end{matrix}\right.\\ \Rightarrow m_{Al}=0,4.27=10,8\left(g\right);m_{Fe}=56.0,2=11,2\left(g\right)\\ n_{HCl}=2.0,8=1,6\left(mol\right)\\ m_{HCl}=1,6.36,5=58,4\left(g\right)\\ m_{ddHCl}=\dfrac{58,4.100}{25}=233,6\left(g\right)\)

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

x           3x             x             1,5x

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

y         2y              y          y

\(\left\{{}\begin{matrix}27x+56y=22\\1,5x+y=\dfrac{17,92}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

\(m_{Al}=0,4\cdot27=10,8g\)

\(m_{Fe}=22-10,8=11,2g\)

\(m_{HCl}=36,5\cdot\left(3x+2y\right)=36,5\cdot\left(3\cdot0,4+2\cdot0,2\right)=58,4g\)

\(m_{ddHCl}=\dfrac{m_{HCl}}{C\%}\cdot100\%=\dfrac{58,4}{25\%}\cdot100\%=233,6g\)

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\)

\(n_{HCl}=0,2\cdot4=0,8mol\)

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

\(x\)   \(\rightarrow\)   \(3x\)            \(x\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 \(y\)   \(\rightarrow\) \(2y\)            \(y\)

\(\Rightarrow\left\{{}\begin{matrix}27x+65y=11,9\\3x+2y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

a)\(\%m_{Al}=\dfrac{0,2\cdot27}{11,9}\cdot100\%=45,38\%\)

\(\%m_{Zn}=100\%-45,38\%=54,62\%\)

b)\(\Sigma n_{H_2}=\dfrac{3}{2}x+y=\dfrac{3}{2}\cdot0,2+0,1=0,4mol\)

\(V_{H_2}=0,4\cdot22.4=8,96l\)

\(Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+27b=5,1\\22,4a+22,4.1,5.b=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ a,\Rightarrow\%m_{Mg}=\dfrac{0,1.24}{5,1}.100\approx47,059\%\\ \Rightarrow\%m_{Al}\approx100\%-47,059\%\approx52,941\%\\ b,n_{HCl}=2.n_{H_2}=2.\left(0,1+0,1.1,5\right)=0,5\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)\)

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

   x           2x            x             x

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

y            3y          y             1,5y

Ta có hệ:

\(\left\{{}\begin{matrix}24x+27y=5,1\\x+1,5y=\dfrac{5,6}{22,4}=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%=47,06\%\)

\(\%m_{Al}=100\%-47,06\%=52,94\%\)

b)\(\Sigma n_{HCl}=2x+3y=2\cdot0,1+3\cdot0,1=0,5mol\)

\(V=\dfrac{n}{C_M}=\dfrac{0,5}{2}=0,25l=250ml\)

a) Gọi số mol Al, Fe là a, b (mol)

=> 27a + 56b = 11,1 (1)

\(n_{HCl}=\dfrac{60.36,5\%}{36,5}=0,6\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            a--->3a-------->a----->1,5a

             Fe + 2HCl --> FeCl₂ + H₂

                b--->2b------->b----->b

=> 3a + 2b = 0,6 (2)

(1)(2) => a = 0,1; b = 0,15 

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{11,1}.100\%=24,32\%\\\%m_{Fe}=\dfrac{0,15.56}{11,1}.100\%=75,68\%\end{matrix}\right.\)

b) \(n_{H_2}=1,5a+b=\) 0,3 (mol)

=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

m_{dd sau pư} = 11,1 + 60 - 0,3.2 = 70,5 (g)

\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,1.133,5}{70,5}.100\%=18,94\%\\C\%_{FeCl_2}=\dfrac{0,15.127}{70,5}.100\%=27,02\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

\(m_{HCl}=\dfrac{60\cdot36,5}{100}=21,9g\)

\(\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\)

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

x          3x                          1,5x

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

y         2y                        y

\(\Rightarrow\left\{{}\begin{matrix}27x+56y=11,1\\3x+2y=0,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)

a)\(\%m_{Al}=\dfrac{0,1\cdot27}{11,1}\cdot100\%=24,32\%\)

\(\%m_{Fe}=100\%-24,32\%=75,68\%\)

b)\(\Sigma n_{H_2}=1,5x+y=1,5\cdot0,1+0,15=0,3mol\)

\(V_{H_2}=0,3\cdot22,4=6,72l\)

a)Gọi x,y lần lượt là số mol của Al, Fe trong hỗn hợp ban đầu (x,y>0)

Sau phản ứng hỗn hợp muối khan gồm: \(\left\{{}\begin{matrix}AlCl_3:x\left(mol\right)\\FeCl_2:y\left(mol\right)\end{matrix}\right.\)

Ta có hệ phương trình: \(\left\{{}\begin{matrix}27x+56y=13,9\\133,5x+127y=38\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\approx0,0896\\y\approx0,205\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,0896\cdot27\cdot100\%}{13,9}\approx17,4\%\\\%m_{Fe}=\dfrac{0,205\cdot56\cdot100\%}{13,9}\approx82,6\%\end{matrix}\right.\)

Theo Bảo toàn nguyên tố Cl, H ta có:\(n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{3n_{AlCl_3}+2n_{FeCl_2}}{2}\\ =\dfrac{3\cdot0,0896+2\cdot0,205}{2}=0,3394mol\\ \Rightarrow V_{H_2}=0,3394\cdot22,4\approx7,6l\)

Theo gt ta có: $n_{H_2}=0,8(mol)$

a, Gọi số mol Fe và Al lần lượt là a;b(mol)

$Fe+2HCl\rightarrow FeCl_2+H_2$

$2Al+6HCl\rightarrow 2AlCl_3+3H_2$

Ta có: $56a+27b=22;a+1,5b=0,8$

Giải hệ ta được $a=0,2;b=0,4$

Do đó $\%m_{Fe}=50,9\%;\%m_{Al}=49,1\%$

b, Sau phản ứng dung dịch chứa 0,2mol $FeCl_2$ và 0,4mol $AlCl_3$

$m_{dd}=22+1,6.36,5:7,3\%-0,8.2=820,4(g)$

Do đó $\%C_{FeCl_2}=3,09\%;\%C_{AlCl_3}=6,5\%$

- Kim loại Cu sẽ không tan trong dung dịch HCl ở đk thường. Nên nó sẽ là kim loại duy nhất trong hỗn hợp này tác dụng với dd H₂SO_{4 đặc,nóng} .

\(Cu+2H_2SO_{4\left(đặc,nóng\right)}\rightarrow CuSO_4+SO_2+H_2O\)

Ta có: \(n_{Cu}=n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

=> m_Cu= 0,1.64=6,4(g)

\(\rightarrow m_{hh\left(Mg,Al\right)}=11,5-6,4=5,1\left(g\right)\\ Đặt\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \rightarrow\left\{{}\begin{matrix}24a+27b=5,1\\a+1,5b=\dfrac{5,6}{22,4}=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \left\{{}\begin{matrix}\%m_{Cu}=\dfrac{6,4}{11,5}.100\approx55,652\%\\\%m_{Mg}=\dfrac{24.0,1}{11,5}.100\approx20,87\%\\\%m_{Al}=\dfrac{27.0,1}{11,5}.100\approx23,478\%\end{matrix}\right.\)

\(a.n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ a............3a.......a.........1,5a\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.........2b........b.........b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}27a+56b=5,5\\1,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,5}.100\approx49,091\%\\\%m_{Fe}=\dfrac{0,05.56}{5,5}.100\approx50,909\%\end{matrix}\right.\\ b.C_{MddHCl}=\dfrac{3a+2b}{0,5}=\dfrac{3.0,1+2.0,05}{0,5}=0,8\left(M\right)\)