\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

_____0,1----->0,15

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)

\(n_{Fe_2O_3}=0,2(mol)\\ Fe_2O_3+6HCl \to 2FeCl_3+3H_2O\\ n_{HCl}=1,2(mol)\\ V_{ddHCl}=\frac{250}{1,25}=200(ml)=0,2(l)\\ CM_{HCl}=\frac{1,2}{0,2}=6M$\)

PTHH: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(n_{H_2SO_4}=3n_{Al_2O_3}=3\cdot\dfrac{15,3}{102}=0,45\left(mol\right)\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,45}{3}=0,15\left(M\right)\)

a) Al2O3 + 3H2SO4 ->Al2(SO4)3 + 3H2O

nAl2O3 = m/M = 15.3/102 = 0.15

CM H2SO4 = n/V = 0.45/3 = 0.15

a)

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$

$V = 0,3.22,4 = 6,72(lít)$
$C_{M_{H_2SO_4}} = \dfrac{0,3}{0,25} = 1,2M$

b)

$n_{CuO} = \dfrac{16}{80} = 0,2(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{CuO} < n_{H_2}$ nên $H_2$ dư

$n_{Cu} = n_{CuO} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$

\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

       0,3     0,3                        0,3 (mol)

a) V= n. 22,4 = 0,3 . 22,4 = 6,72(l)

 \(C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%=\dfrac{0,3.98}{200}.100\%=11,76\%\)

b) PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

         0,3                0,3 

=> m_Cu=n.M=0,3.64=19,2(g)

a) 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)

Ta có: \(n_{Fe_2O_3}=\dfrac{9,6}{160}=0,06\left(mol\right)\)

a. PTHH: Fe₂O₃ + 3H₂SO₄ ---> Fe₂(SO₄)₃ + 3H₂O (1)

Theo PT₍₁₎: \(n_{H_2SO_4}=3.n_{Fe_2O_3}=3.0,06=0,18\left(mol\right)\)

=> \(m_{H_2SO_4}=0,18.98=17,64\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{17,64}{m_{dd_{H_2SO_4}}}.100\%=9,8\%\)

=> \(m_{dd_{H_2SO_4}}=180\left(g\right)\)

b. Ta có: \(m_{dd_{Fe_2\left(SO_4\right)_3}}=9,6+180=189,6\left(g\right)\)

Theo PT₍₁₎: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,06\left(mol\right)\)

=> \(m_{Fe_2\left(SO_4\right)_3}=0,06.400=24\left(g\right)\)

=> \(C_{\%_{Fe_2\left(SO_4\right)_3}}=\dfrac{24}{189,6}.100\%=12,66\%\)

c. PTHH: Fe₂(SO₄)₃ + 3BaCl₂ ---> 3BaSO₄↓ + 2FeCl₃ (2)

Theo PT₍₂₎: \(n_{BaSO_4}=3.n_{Fe_2\left(SO_4\right)_3}=3.0,06=0,18\left(mol\right)\)

=> \(m_{BaSO_4}=0,18.233=41,94\left(g\right)\)

Theo PT₍₂₎: \(n_{BaCl_2}=n_{BaSO_4}=0,18\left(mol\right)\)

=> \(m_{BaCl_2}=0,18.208=37,44\left(g\right)\)

Ta có: \(C_{\%_{BaCl_2}}=\dfrac{37,44}{m_{dd_{BaCl_2}}}.100\%=10,4\%\)

=> \(m_{dd_{BaCl_2}}=360\left(g\right)\)