a, Ta có: 27n_Al + 56n_Fe = 22 (1)

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

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)

b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)

a, Ta có: 27n_Al + 56n_Fe = 27,8 (1)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{17,353}{24,79}=0,7\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{27,8}.100\%\approx19,42\%\\\%m_{Fe}\approx80,58\%\end{matrix}\right.\)

b, \(n_{H_2SO_4}=n_{H_2}=0,7\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,7}{0,5}=1,4\left(M\right)\)

• pt: Zn + 2HCl -> ZnCl₂ +H₂ 
• n_HCl = ( 3,25 : 65 ) x 2 = 0,1 (mol)

V = 0,1 : 0,5 = 0,2 (l)

• gọi a là số mol cần tìm
• pt: 2Al + 3H₂SO₄ -> Al₂(SO4)₃ + 3H₂

_​                 a                       ->                 3/2a

Fe  + H₂SO₄  -> FeSO₄  + H₂

a                        ->             a

• ta có : a + 3/2a = 0,05  => a = 0,02 (mol)
• C%_Fe = ( 0,02 x 56)x100 / (0,02x56 + 0,02x 27) = 67,47%
• C% _Al = 100 -67,47= 32,53%

c

a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b) Ta có: \(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)=n_{Zn}\)

\(\Rightarrow m_{Zn}=65\cdot0,1=6,5\left(g\right)\) \(\Rightarrow m_{Cu}=3,5\left(g\right)\)

c) Theo PTHH: \(n_{HCl}=2n_{Zn}=0,2mol\)

\(\Rightarrow V_{HCl}=\frac{0,2}{0,2}=1\left(l\right)\)

HCl cơ mà !!

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{H_2}=0,2(mol)\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,1}=2M\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow m_{Cu}=20-5,6=14,4(g)\\ c,\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\ \%m_{Cu}=100\%-28\%=72\%\)