==0,06 (mol)

 ko pứ

0,06    0,12                ←0,06          (mol)

%=.100 % ≈25,85 %

%=100 % - 25,85 %=74,15 %

///

=0,12.26,5=4,38 (g)

= =29,2 (g)

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

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

\(0.1.......0.2......................0.1\)

Chất rắn X : Cu 

\(m_{Zn}=0.1\cdot65=6.5\left(g\right)\Rightarrow m_{Cu}=19.3-6.5=12.8\left(g\right)\)

\(n_{Cu}=\dfrac{12.8}{64}=0.2\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

\(2Cu+O_2\underrightarrow{^{^{t^o}}}2CuO\)

\(0.2........0.1\)

\(m_{tăng}=m_{O_2}=0.1\cdot32=3.2\left(g\right)\)

a. B gồm AlCl₃ 

\(b.n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,4      1,2             0,4           0,6

\(\%m_{Al}=\dfrac{0,4.27}{23,6}\cdot100\%=45,76\%\\ \%m_{Cu}=100\%-45,76=54,24\%\\ c.m_B=m_{AlCl_3}=1,2.133,5=106,2g\)

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 % 

\(Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}24x+27y=10,2\\x+1,5y=0,5\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,2.24}{10,2}.100=47,06\%\\ \%m_{Al}=52,94\%\\ n_{HCl}=2n_{Mg}+3n_{Al}=0,2.2+0,3.2=1\left(mol\right)\\ \Rightarrow V_{HCl}=\dfrac{1}{2}=0,5\left(l\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,04\left(mol\right)\Rightarrow m_{Zn}=0,04.65=2,6\left(g\right)\)

⇒ m_Cu = 9 - 2,6 = 6,4 (g)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{2,6}{9}.100\%\approx28,89\%\\\%m_{Cu}\approx71,11\%\end{matrix}\right.\)