\(a.2Al+6HCl->2AlCl_3+3H_2\\ Mg+2HCl->MgCl_2+H_2\\ b.n_{Al}=a,n_{Mg}=b\\ 27a+24b=7,5\left(I\right)\\ 1,5a+b=\dfrac{7,84}{22,4}=0,35\left(II\right)\\ a=0,1;b=0,2\\ \%m_{Al}=\dfrac{27\cdot0,1}{7,5}\cdot100\%=36\%\\ \%m_{Mg}=64\%\\ c.m_{HCl}=36,5\left(0,1\cdot3+0,2\cdot2\right)=18,25g\\ d.m_{ddsau}=7,5+\dfrac{18,25}{14,6:100}-0,35\cdot2=131,8g\\ C\%\left(AlCl_3\right)=\dfrac{133,5\cdot0,1}{131,8}\cdot100\%=10,1\%\\ C\%\left(MgCl_2\right)=\dfrac{95\cdot0,2}{131,8}\cdot100\%=14,4\%\)

a,\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:      x                                 x    

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:      y                                 y

Ta có: \(\left\{{}\begin{matrix}65x+56y=30,7\\x+y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{Zn}=\dfrac{0,3.65.100\%}{30,7}=63,52\%;\%m_{Fe}=100\%-63,52\%=36,48\%\)

b,

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,3     0,6                             

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,2     0,4

n_HCl = 0,6+0,4 = 1 (mol)

\(V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)\)

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{20,16}{22,4}=0,9mol\)

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Mg}=y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=27x\\m_{Mg}=24y\end{matrix}\right.\)

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

 x                                      \(\dfrac{3}{2}x\)    ( mol )

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

  y                                     y ( mol )

Ta có:

\(\left\{{}\begin{matrix}27x+24y=19,8\\\dfrac{3}{2}x+y=0,9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,6\end{matrix}\right.\)

\(\Rightarrow m_{Al}=0,2.27=5,4g\)

\(\Rightarrow m_{Mg}=0,6.24=14,4g\)

\(\%m_{Al}=\dfrac{5,4}{19,8}.100=27,27\%\)

\(\%m_{Mg}=100\%-27,27\%=72,73\%\)

- Xét TN2:

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 2NaOH + 2H₂O --> 2NaAlO₂ + 3H₂

            0,2<--------------------------------0,3

=> m_Al = 0,2.27 = 5,4 (g)

- Xét TN1:

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

           0,2------------------->0,3

             Mg + 2HCl --> MgCl₂ + H₂

          0,15<------------------0,15

=> m_Mg = 0,15.24 = 3,6 (g)

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

Lập hệ phương trình ( Al là x , Mg là y )

\(\left\{{}\begin{matrix}27x+24y=15\\27x=\dfrac{36}{100}.15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)

2Al + 6HCl ---> 2AlCl3 + 3H2

0,2                                    0,3

2Mg + 4HCl ---> 2MgCl2 + 2H2

0,4                                   0,4

\(\Sigma n_{H2\uparrow}=0,3+0,4=0,7\left(mol\right)\)

\(\Rightarrow V_{H2\uparrow}=0,7.22,4=15,68\left(l\right)\)