Ta có nH2 = 3,36/22,4 = 0,15 mol
        Fe +2 HCl -> FeCl2 + H2
       0,15.    0,3                   <-. 0,15.  ( Mol)
=> mFe = 0,15 × 56 = 8,4g
   => %Fe = 8,4/15×100% = 56% 
      => %Cu = 100% - 56% = 44%

=>VHCl =1\0,3=10\3 l

PTHH :  2Fe    +     6HCl   -->   2FeCl₃  +  3H₂   (1)

n_H2 =  \(\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

Từ (1) ->  n_Fe = \(\dfrac{2}{3}n_{H_2}=0.1\left(mol\right)\)

-> m_Fe = n.M = 0,1  . 56 = 5.6 (g) => %m_Fe = \(\dfrac{5.6}{15}x100\%\approx37.3\%\)

-> %m_Cu =  100% - 37.3% = 62.7 % 

Ta có: \(\left\{{}\begin{matrix}x=Fe\\y=Cu\end{matrix}\right.\) trong 40g hh

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

PTHH: Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

  TL:      1        2            1         1

mol:     0,5     \(\leftarrow\)  1    \(\leftarrow\)   0,5  \(\leftarrow\) 0,5

\(m_{Fe}=n.M=0,5.56=28g\)

\(\%m_{Fe}=\dfrac{m_{Fe}}{m_{hh}}.100\%=\dfrac{28}{40}.100\%=70\%\)

\(\%m_{Cu}=100\%-70\%=30\%\)

\(m_{không.tan}=m_{Cu}=2\left(g\right)\\ \Rightarrow m_{Al,Fe}=10,3-2=8,3\left(g\right)\\ Đặt:a=n_{Al}\left(mol\right);b=n_{Fe}\left(mol\right)\left(a,b>0\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=8,3\\1,5.22,4a+22,4b=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{2}{10,3}.100\approx19,417\%\\ \%m_{Fe}=\dfrac{56.0,1}{10,3}.100\approx54,369\%\\ \Rightarrow\%m_{Al}\approx26,214\%\)

1)

Fe + 2HCl --> FeCl₂ + H₂

Cu + 2H₂SO₄ --> CuSO₄ + SO₂ + 2H₂O

2)

- Xét TN1:

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

         0,15<------------------0,15

=> m_Fe = 0,15.56 = 8,4 (g)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{8,4}{14,8}.100\%=56,757\%\\\%m_{Cu}=100\%-56,757\%=43,243\%\end{matrix}\right.\)

3) 

- Xét TN2:

\(n_{Cu}=\dfrac{29,6.43,243\%}{64}=0,2\left(mol\right)\)

PTHH: Cu + 2H₂SO₄ --> CuSO₄ + SO₂ + 2H₂O

            0,2-------------------------->0,2

=> V = 0,2.22,4 = 4,48 (l)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=1\left(mol\right)\Rightarrow m_{Zn}=1.65=65\left(g\right)\)

\(\Rightarrow m_{Cu}=80,5-65=15,5\left(g\right)\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,15<-0,3<----0,15<---0,15

\(\left\{{}\begin{matrix}\%Fe=\dfrac{0,15.56}{12}.100\%=70\%\%\\\%Cu=100\%-70\%=30\%\end{matrix}\right.\)

c) m_HCl = 0,3.36,5 = 10,95 (g)

=> \(m_{dd}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)

d) mdd  = 12 + 109,5 - 0,15.2 = 121,2 (g)

 \(C\%\left(FeCl_2\right)=\dfrac{0,15.127}{121,2}.100\%=15,718\%\)

a: \(Cu+2HCl->CuCl_2+H_2\)

\(Fe+2HCl->FeCl_2+H_2\)