a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ 2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)

Cho hỗn hợp tác dụng với NaOH, chất rắn không tan là Fe

=> m_Fe= 1,12 (g) \(\Rightarrow n_{Fe}=0,02\left(mol\right)\)

Ta có: \(n_{H_2\left(2\right)}=n_{Fe}=0,02\left(mol\right)\)

=> \(n_{H_2\left(1\right)}=\Sigma n_{H_2}-n_{H_2\left(2\right)}=0,065-0,02=0,045\left(mol\right)\)

\(\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2\left(1\right)}=0,03\left(mol\right)\)

\(\Rightarrow m_{Al}=0,03.27=0,81\left(g\right)\)

\(\Rightarrow\%m_{Al}=41,97\%,\%m_{Fe}=58,03\%\)

b) \(m_{FeCl_2}=0,02.127=2,54\left(g\right)\\ m_{AlCl_3}=0,03.133,5=4,005\left(g\right)\)

Vì Ag không tác dụng với H₂SO₄ loãng 

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

         2           3                    1              3

       0,3                                               0,45

\a) Chất rắn không tan là Ag nên : 

\(m_{Ag}=5,4\left(g\right)\)

⇒ \(m_{Al}=13,5-5,4=8,1\left(g\right)\)

⁰/₀Al = \(\dfrac{8,1.100}{13,5}=60\)⁰/₀

⁰/₀Ag = \(\dfrac{5,4.100}{13,5}=40\)⁰/₀

b) Có : \(m_{Al}=8,1\left(g\right)\)

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

\(n_{H2}=\dfrac{0,3.3}{2}=0,45\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,45.22,4=10,08\left(l\right)\)

 Chúc bạn học tốt

Phần 1 : 

$m_{Cu} = 0,4(gam)$

Gọi $n_{Fe} = a ; n_{Al} = b \Rightarrow 56a + 27b + 0,4 = 1,5 : 2 = 0,75(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{H_2} = a + 1,5b = \dfrac{896}{1000.22,4} = 0,04(2)$

Từ (1)(2) suy ra a = -0,025 < 0$

$\to$ Sai đề

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,25.56=14\left(g\right)\)

m_Cu = 20,4 - 14 = 6,4 (g)

b, \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{14}{20,4}.100\%\approx68,63\%\\\%m_{Cu}\approx31,37\%\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{0,5.36,5}{200}.100\%=9,125\%\)

a) Fe + 2HCl --> FeCl₂ + H₂

FeO + 2HCl --> FeCl₂ + H₂O

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

______0,5<-1<------0,5<---0,5

=> m_Fe = 0,5.56 = 28 (g)

=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{28}{100}.100\%=28\%\\\%FeO=100\%-28\%=72\%\end{matrix}\right.\)

c) \(n_{FeO}=\dfrac{72}{72}=1\left(mol\right)\)

PTHH: FeO + 2HCl --> FeCl₂ + H₂O

______1---->2

=> m_HCl = (1+2).36,5 = 109,5 (g)

=> \(m_{ddHCl}=\dfrac{109,5.100}{30}=365\left(g\right)\)

=> \(V_{ddHCl}=\dfrac{365}{1,15}=317,39\left(ml\right)\)

Chất rắn không tan là Vàng $\Rightarrow m_{Au} = 1,97(gam)$

Gọi $n_{Cu} = a(mol) ; n_{Ag} = b(mol) \Rightarrow 64a + 108b = 6,05 - 1,97(1)$

$n_{NO_2} = \dfrac{1,792}{22,4} = 0,08(mol)$

Bảo toàn e : $2a + b = 0,08(2)$

Từ (1)(2) suy ra : a = 0,03 ; b = 0,02

$\%m_{Au} = \dfrac{1,97}{6,05}.100\% = 32,6\%$

$\%m_{Cu} = \dfrac{0,03.64}{6,05}.100\% = 31,7\%$

$\%m_{Ag} = 100\% - 32,6\% - 31,7\% = 36,7\%$

Theo bài ra, ta có: \(m_{Ag}=5,6\left(g\right)\)

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

a) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Al}=\dfrac{1}{15}\left(mol\right)\) \(\Rightarrow m_{Al}=\dfrac{1}{15}\cdot27=1,8\left(g\right)\)

\(\Rightarrow\%m_{Al}=\dfrac{1,8}{1,8+5,6}\cdot100\%\approx24,32\%\) \(\Rightarrow\%m_{Ag}=75,68\%\)

b) Theo PTHH: \(n_{H_2SO_4}=n_{H_2}=0,1mol\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

c) PTHH: \(H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+2H_2O\)

Theo PTHH: \(n_{Ba\left(OH\right)_2}=n_{H_2SO_4}=0,1mol\)

\(\Rightarrow V_{ddBa\left(OH\right)_2}=\dfrac{0,1}{0,2}=0,5\left(l\right)=500\left(ml\right)\)