Gọi x là số mol của Al, Zn

n_H2 = \(\dfrac{13,44}{22,4}=0,6\) mol

Pt: 2Al + 6HCl --> 2AlCl₃ + 3H₂

......x........................................1,5x

.....Zn + 2HCl --> ZnCl₂ + H₂

......x....................................x

Ta có: 1,5x + x = 0,6

=> x = 0,24

m_hh = 0,24 . (27 + 65) = 22,08 (g)

Theo pt ta có: n_HCl = 2n_H2 = 2 . 0,6 = 1,2 mol

m_HCl = 1,2 . 36,5 = 43,8 (g)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)

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

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

Có \(\Sigma n_{H_2}=n_{Mg}+n_{Zn}=0,6\)

Mà \(n_{Mg}=n_{Zn}\Rightarrow n_{Mg}=n_{Zn}=0,3mol\)

\(m_{Mg}=0,3\cdot24=7,2g\)

\(m_{Zn}=0,3\cdot65=19,5g\)

\(\Sigma n_{HCl}=2n_{Mg}+2n_{Zn}=2\cdot0,3+2\cdot0,3=1,2mol\)

\(\Rightarrow m_{HCl}=1,2\cdot36,5=43,8g\)

Gọi n_Al = n_Zn = a (mol)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

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

             Zn + 2HCl --> ZnCl₂ + H₂

             a--------------------->a

=> 1,5a + a = 0,6

=> a = 0,24 (mol)

=> m_hh = 0,24.27 + 0,24.65 = 22,08 (g)

Gọi \(n_{Al}=n_{Zn}=a\left(mol\right)\)

PTHH:

2Al + 6HCl ---> 2AlCl₃ + 3H₂

a                                       a

Zn + 2HCl ---> ZnCl₂ + H₂

a                                     a

\(\rightarrow22,4\left(a+a\right)=13,44\\ \Leftrightarrow a=0,3\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}m_{Al}=0,3.27=8,1\left(g\right)\\m_{Zn}=0,3.65=19,5\left(g\right)\end{matrix}\right.\\ \rightarrow m_{hh}=8,1+19,5=27,6\left(g\right)\)

*Sửa đề: "13,44 lít H₂" và "24,9 gam hh 2 kim loại"

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

               a_____3a_____________\(\dfrac{3}{2}\)a                (mol)

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

               b_____2b_____________b                   (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}27a+65b=24,9\\\dfrac{3}{2}a+b=\dfrac{13,44}{22,4}=0,6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Zn}=0,3\left(mol\right)\\n_{HCl}=1,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2\cdot27=5,4\left(g\right)\\m_{Zn}=19,5\left(g\right)\\m_{ddHCl}=\dfrac{1,2\cdot36,5}{7,3\%}=600\left(g\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}Fe\\Al\end{matrix}\right.+HCl->\left\{{}\begin{matrix}FeCl2\\AlCl3\end{matrix}\right.+H2\)

Ta có số mol Fe là x , Al là y (mol)

\(\left\{{}\begin{matrix}56x+27y=11\\127x+133,5y=39,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%mFe=\dfrac{0,1.56}{11}=50,9\%\\\%mAl=\dfrac{0,2.27}{11}=49,09\%\end{matrix}\right.\)

Bảo toàn e :

\(2.nH2=2.nFe+3.nAl\Rightarrow nH2=0,4\left(mol\right)\)

\(V=0,4.22,4=8,96\left(l\right)\)

\(nFe=nFeCl2=0,1\left(mol\right)\)

\(nAl=nAlCl3=0,2\left(mol\right)\)

\(\Rightarrow nHCl\left(pứ\right)=2.0,1+3.0,2=0,8\left(mol\right)\)

\(Cm=\dfrac{n}{V}=\dfrac{0,8}{0,2}=4\left(M\right)\)