\(Cu+2H_2SO_{4\left(đ,n\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

\(MgSO_3+H_2SO_4\rightarrow MgSO_4+SO_2+H_2O\)

\(S+2H_2SO_{4\left(đ,n\right)}\rightarrow3SO_2+2H_2O\)

\(2Al+6H_2SO_{4\left(đ,n\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

\(2FeO+4H_2SO_{4\left(đ,n\right)}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)

\(Ba\left(NO_3\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HNO_3\)

\(BaO+H_2SO_4\rightarrow BaSO_4+H_2O\)

Bảo toàn điện tích: x + 2y = 0,8 (1)

Ta có: 56.0,1 + 27.0,2 + 35,5x + 96y = 46,9

=> 35,5x + 96y = 35,9 (2)

(1)(2) => x = 0,2 (mol); y = 0,3 (mol)

a) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,05}{1}\) => Fe dư, H₂SO₄ hết

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

         0,05<---0,05------>0,05--->0,05

=> V = 0,05.22,4 = 1,12 (l)

b) 

PTHH: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

                          0,05->0,0375

=> m_Fe = 0,0375.56 = 2,1 (g)

a)

CTHH: M₂O₃

\(n_{M_2O_3}=\dfrac{4}{2.M_M+48}\left(mol\right)\)

PTHH: \(M_2O_3+3H_2SO_4\rightarrow M_2\left(SO_4\right)_3+3H_2O\)

Theo PTHH: \(n_{H_2SO_4}=3.n_{M_2O_3}=\dfrac{12}{2.M_M+48}\left(mol\right)\)

=> \(m_{H_2SO_4}=\dfrac{588}{M_M+24}\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{\dfrac{588}{M_M+24}.100}{9,8}=\dfrac{6000}{M_M+24}\left(g\right)\)

m_{dd sau pư} = \(4+\dfrac{6000}{M_M+24}\left(g\right)\)

Theo PTHH: \(n_{M_2\left(SO_4\right)_3}=n_{M_2O_3}=\dfrac{4}{2.M_M+48}\left(mol\right)\)

Ta có: \(C\%_{M_2\left(SO_4\right)_3}=\dfrac{\dfrac{4}{2.M_M+48}\left(2.M_M+288\right)}{4+\dfrac{6000}{M_M+24}}.100\%=12,658\%\)

=> M_M = 56 (g/mol)

=> M là Fe

CTHH của oxit là Fe₂O₃

\(m_{dd.H_2SO_4}=\) 75 (g)

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

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

          0,05--->0,05----->0,05--->0,05

\(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{4,9}{250}.100\%=1,96\%\)

\(V_{dd.sau.pư}=\dfrac{0,05}{2}=0,025\left(l\right)\)

\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

a) 

\(n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\)

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

          0,15-->0,45----->0,15--->0,225

V = 0,225.22,4 = 5,04 (l)

b) \(m_{HCl}=0,45.36,5=16,425\left(g\right)\Rightarrow m_{dd.HCl}=\dfrac{16,425.100}{10}=164,25\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{50.4,9\%}{98}=0,025\left(mol\right)\)

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,025}{3}\) => Fe₂O₃ dư, H₂SO₄ hết

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

            \(\dfrac{0,025}{3}\)<--0,025------>\(\dfrac{0,025}{3}\)

\(m_A=\dfrac{0,025}{3}.160+50=\dfrac{154}{3}\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{\dfrac{0,025}{3}.400}{\dfrac{154}{3}}.100\%=6,5\%\)

Phần 1:

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

           0,2<---0,4<-------------0,2

=> m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{14,6}=100\left(g\right)\)

Phần 2:

\(n_{SO_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PTHH: \(2Fe+6H_2SO_{4\left(đ,n\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

             0,2----------------------------------->0,3

            \(Cu+2H_2SO_{4\left(đ,n\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

           0,15<------------------------------0,15

=> m_X = 2.(0,2.56 + 0,15.64) = 41,6 (g)

bn check lại xem đề sai đâu k nhé

Gọi số mol Al, Fe là a, b (mol)

=> 27a + 56b = 11 (1)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

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

               a---->3a------->a----->1,5a

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

              b----->2b------>b----->b

=> 1,5a + b = 0,4 (2)

(1)(2) => a = 0,2 (mol); b = 0,1 (mol)
\(\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

n_HCl = 2a + 2b = 0,8 (mol)

=> m_HCl = 0,8.36,5 = 29,2 (g)

=> \(m_{dd.HCl}=\dfrac{29,2.100}{3,65}=800\left(g\right)\)

m_{dd sau pư} = 800 + 11 - 0,4.2 = 810,2 (g)

\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{810,2}.100\%=3,3\%\\C\%_{FeCl_2}=\dfrac{0,1.127}{810,2}.100\%=1,57\%\end{matrix}\right.\)