a) 

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

b)

\(Ba\left(OH\right)_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+Ag_2O+H_2O\\ Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

\(a,n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=0,2mol\\ m_{Fe}=0,2.56=11,2g\\ m_{Cu}=25-11,2=13,8g\\ b,\%m_{Fe}=\dfrac{11,2}{25}\cdot100=44,8\%\\ \%m_{Cu}=100-44,8=55,2\%\)
c, Gọi CTHH của sắt là \(Fe_xO_y\)

\(Fe_xO_y+yH_2\xrightarrow[t^0]{}xFe+yH_2O\\ \Rightarrow n_{Fe_xO_y}=n_{H_2}:y\\ \Leftrightarrow\dfrac{11,6}{56x+16y}=\dfrac{0,2}{y}\\ \Leftrightarrow11,6y=11,2x+3,2y\\ \Leftrightarrow11,6y-3,2y=11,2x\\ \Leftrightarrow8,4y=11,2x\\ \Leftrightarrow\dfrac{x}{y}=\dfrac{8,4}{11,2}=\dfrac{3}{4}\\ \Rightarrow x=3;y=4\\ \Rightarrow CTHH:Fe_3O_4\)

Câu 1

\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\\ n_{O_2}=\dfrac{2,479}{24,79}=0,1mol\\ 3Fe+2O_2\xrightarrow[t^0]{}Fe_3O_4\\ \Rightarrow\dfrac{0,3}{3}>\dfrac{0,1}{2}\Rightarrow Fe.dư\\ 3Fe+2O_2\xrightarrow[t^0]{}Fe_3O_4\)

0,15       0,1         0,05

\(m_{Fe.dư}=16,8-0,15.56=8,4g\\ b.m_{Fe_3O_4}=0,05.232=11,6g\)

Câu 2

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ R+H_2SO_4\rightarrow RSO_4+H_2\\ n_{H_2}=n_R=0,3mol\\ M_R=\dfrac{12}{0,3}=40g/mol\)

Vậy M là Canxi

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{Zn}=n_{H_2}=0,1mol\\ \%m_{Zn}=\dfrac{0,1.65}{10,5}\cdot100=61,9\%\\ \%m_{Cu}=100-61,9=38,1\%\)

A.

B

\(\left(1\right)2Fe+3Cl_2\xrightarrow[t^0]{}2FeCl_3\\ \left(2\right)FeCl_2+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\\ \left(3\right)2Fe\left(OH\right)_3\xrightarrow[t^0]{}Fe_2O_3+3H_2O\\ \left(4\right)Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

____________

\(\left(1\right)2Al+3Cl_2\rightarrow2AlCl_3\\ \left(2\right)AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\\ \left(3\right)Al\left(NO_3\right)_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaNO_3\\ \left(4\right)2Al\left(OH\right)_3\xrightarrow[t^0]{}Al_2O_3+3H_2O\\ \left(5\right)Al_2O_3+3CO\xrightarrow[]{t^0}2Al+3CO_2\\ \left(6\right)4Al+3O_2\xrightarrow[]{t^0}2Al_2O_3\\ \left(7\right)2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\\ \left(8\right)Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ 2FeCl_2+Cl_2\xrightarrow[]{t^0}2FeCl_3\\ FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\\ 2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\\ Fe_2\left(SO_4\right)_3+3BaCl_2\rightarrow2FeCl_2+3BaSO_4\\ 2FeCl_2+Fe\xrightarrow[]{t^0}3FeCl_2\\ FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ FeCl_2+Ag_2SO_4\rightarrow FeSO_4+2AgCl\\ FeSO_4+Ba\left(OH\right)_2\rightarrow Fe\left(OH\right)_2+BaSO_4\\ Fe\left(OH\right)_2\xrightarrow[]{t^0}FeO+H_2O\\ FeO+H_2\xrightarrow[]{t^0}Fe+H_2O\)