a)\(3M+4nHNO_3-->3M\left(NO_3\right)_n+nNO+2nH_2O\)

b)

\(2M+2nH_2SO_4-->M_2\left(SO_4\right)_n+nSO_2+2nH_2O\)

c)

\(8M+30HNO_3-->8M\left(NO_3\right)_3+3N_2O+15H_2O\)

d)

\(8M+10nHNO_3-->8M\left(NO_3\right)_n+nN_2O+5nH_2O\)

e)\(\left(5x-2y\right)Fe+\left(15x-3y\right)HNO_3-->\left(5x-2y\right)Fe\left(NO_3\right)_3+3N_xO_y+\left(\dfrac{15x-3y}{2}\right)H_2O\)

f) \(3Fe_xO_y+\left(6x+2y\right)HNO_3-->3xFe\left(NO_3\right)_3+\left(2y-3x\right)NO+\left(3x+y\right)H_2O\)

g)\(Fe_xO_y+\left(6x-2y\right)HNO_3-->xFe\left(NO_3\right)_3+\left(3x-2y\right)NO_2+\left(3x-y\right)H_2O\) h)\(Fe_xO_y+2yHCl-->xFeCl_{\dfrac{2y}{x}}+yH_2O\)

i)\(2Fe_xO_y+2yH_2SO_4-->xFe_2\left(SO_4\right)_{\dfrac{2y}{x}}+2yH_2O\)

\(a,4Mg+10HNO_3\to 4Mg(NO_3)_2+NH_4NO_3\uparrow+3H_2O\\ b,2Fe_xO_y+(6x-2y)H_2SO_4\to xFe_2(SO_4)_3+(3x-2y)SO_2\uparrow+(6x-2y)H_2O\\ c,3Fe_xO_y+(12x-2y)HNO_3\to 3xFe(NO_3)_3+(3x-2y)NO\uparrow+(6x-y)H_2O\\ d,2FeO+4H_2SO_4\to Fe_2(SO_4)_3+SO_2\uparrow+4H_2O\\ e,3Fe_3O_4+28HNO_3\to 9Fe(NO_3)_3+NO\uparrow+14H_2O\)

Bạn xem lại PT 1 và 3 nhé.

\(\overset{0}{Al}+H\overset{+5}{N}O_3\rightarrow\overset{+3}{Al}\left(NO_3\right)_3+\overset{0}{N_2}+H_2O\)

\(\overset{0}{Al\rightarrow}\overset{+3}{Al}+3e|\times10\)

\(2\overset{+5}{N}+10e\rightarrow\overset{0}{N_2}|\times3\)

⇒ 10Al + 36HNO₃ → 10Al(NO₃)₃ + 3N₂ + 18H₂O

\(\overset{^{+2y/x}}{Fe_x}O_y+H\overset{+5}{N}O_3\rightarrow\overset{+3}{Fe}\left(NO_3\right)_3+\overset{+4}{N}O_2+H_2O\)

\(\overset{^{+2y/x}}{Fe_x}\rightarrow x\overset{+3}{Fe}+\left(3x-2y\right)e|\times1\)

\(\overset{+5}{N}+e\rightarrow\overset{+4}{N}|\times\left(3x-2y\right)\)

⇒ Fe_xO_y + (6x-2y)HNO₃ → xFe(NO₃)₃ + (3x-2y)NO₂ + (3x-y)H₂O

1) FeS₂ + 5HNO₃ + 3HCl → FeCl₃ + 2H₂SO₄ + 5NO + 2H₂O (Câu 1 nên sửa Fe₂O₃ thành FeS₂ thì mới viết được phương trình)

2) (5x-2y)Fe + (18x-6y)HNO₃ → (5x-2y)Fe(NO₃)₃ + 3NxOy + (9x-3y)H₂O

3) 3FexOy + (12x-2y)HNO₃ → 3xFe(NO₃)₃ + (3x-2y)NO + (6x-y)H₂O

4) 2FexOy + (6x-2y)H₂SO₄ → xFe₂(SO₄)₃ + (3x-2y)SO₂+ (6x-2y)H₂O

5) 2CrI₃ + 64KOH + 27Cl₂ → 2K₂CrO₄ + 6KIO₄ + 54KCl + 32H₂O

bạn có bị nhầm k vậy? ở phản ứng đầu không có chât tham gia nào mang nguyên tố S cả thê nhưng sau phản ứng co axit sunfuric là sao vậy????

2) \(\left(5x-2y\right)Fe+\left(18-6y\right)HNO_3\rightarrow\left(5x-2y\right)Fe\left(NO_3\right)_3+3N_xO_y+\left(9x-3y\right)H_2O\)

3FexOy + (12x-2y)HNO3 ----> 3xFe(NO3)3 + (3x-2y)NO + (6x-y)H2O

Quá trình:

\(Fe_x^{^{+\dfrac{2y}{x}}}\rightarrow xFe^{+3}+\left(3x-2y\right)|\times4\)

\(N^{+5}+3e\rightarrow N^{+2}|\times\left(3x-2y\right)\) 

\(N^{+5}+1e\rightarrow N^{+4}|\times\left(3x-2y\right)\)

PT: \(4Fe_xO_y+\left(18x-4y\right)HNO_3\rightarrow4xFe\left(NO_3\right)_3+\left(3x-2y\right)NO+\left(3x-2y\right)NO_2+\left(9x-2y\right)H_2O\)

3Fe_xO_y + (12x - 2y)HNO₃ ===> 3xFe(NO₃)₃ + (3x - 2y)NO + (6x - y)H₂O

3Fe_xO_y + (12x - 2y ) HNO₃ => 3xFe(NO₃)₃ + (3x - 2y)NO + (6x - y)H₂O

2Fe(OH)2 + 4H2SO4 ---> Fe2(SO4)3 + SO2 + 6H2O
4Cu2O + 18HNO3 ---> 8Cu(NO3)2 + NH4NO3 + 7H2O
4FexOy + (12x-4y) H2SO4 ---> 2x Fe2(SO4)3 + (6x-4y) SO2 + (12x-4y) H2O
FexOy + (4x-2y) HNO3 ---> x Fe(NO3)2 + (2x-2y) NO2 + (2x-y) H2O
(5t-2z) FexOy + (18xt - 2yt - 6xz) HNO3---> (5tx-2zx) Fe(NO3)3 + (3x-2y) NtOz + (9xt-yt-3xz) H2O

3FexOy + (12x-2y)HNO3 3xFe(NO3)3 + (3x-2y)NO + (6x-y)H2O

3MxOy + (4nx - 2y)HNO3 = 3xM(NO3)n + (nx - 2y)NO + (2nx - y)H2O