Trong phản ứng FexOy + HNO3 ® Fe(NO3)3 + N2 + H2O thì một phân tử FexOy sẽ:
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mk làm ví dụ 3 ý đầu nhé mấy ý sau tương tự
MnO2+HCl\(\rightarrow\)MnCl2+Cl2+H2O
Mn+4 +2e\(\rightarrow\)Mn+2_____.1
2Cl- \(\rightarrow\)Cl2+2e________.1
\(\rightarrow\)MnO2+4HCl\(\rightarrow\)MnCl2+Cl2+2H2O
Al\(\rightarrow\)Al+3 +3e ______.8
N+5 +8e\(\rightarrow\)N-3______.3
\(\rightarrow\)8Al+30HNO3\(\rightarrow\)8Al(NO3)3+3NH4NO3+15H2O
6Fe+\(\frac{8}{3}\rightarrow\)6Fe+3 +2e ___________.3
S+6 +6e\(\rightarrow\)S0 ________________ .1
\(\rightarrow\)6Fe3O4+28H2SO4\(\rightarrow\)9Fe2(SO4)3+S+28H2O
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,4Mg+10HNO_3\to 4Mg(NO_3)_2+NH_4NO_3\uparrow+3H_2O\\ b,2Fe_xO_y+(6x-2y)H_2SO_4\to xFe_2(SO_4)_3+(3x-2y)SO_2\uparrow+(6x-2y)H_2O\\ c,3Fe_xO_y+(12x-2y)HNO_3\to 3xFe(NO_3)_3+(3x-2y)NO\uparrow+(6x-y)H_2O\\ d,2FeO+4H_2SO_4\to Fe_2(SO_4)_3+SO_2\uparrow+4H_2O\\ e,3Fe_3O_4+28HNO_3\to 9Fe(NO_3)_3+NO\uparrow+14H_2O\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\overset{^{+2y/x}}{Fe_x}O_y+H\overset{+5}{N}O_3\rightarrow\overset{+3}{Fe}\left(NO_3\right)_3+\overset{+4}{N}O_2+H_2O\)
\(\overset{^{+2y/x}}{Fe_x}\rightarrow x\overset{+3}{Fe}+\left(3x-2y\right)e|\times1\)
\(\overset{+5}{N}+e\rightarrow\overset{+4}{N}|\times\left(3x-2y\right)\)
⇒ FexOy + (6x-2y)HNO3 → xFe(NO3)3 + (3x-2y)NO2 + (3x-y)H2O
![](https://rs.olm.vn/images/avt/0.png?1311)
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)\(2Fe_xO_y+2yH_2SO_4-->xFe_2\left(SO_4\right)_{\dfrac{2y}{x}}+2yH_2O\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án C.
(5x-2y)Fe3O4 + (46x -18y)HNO3 ® (15x- 6y)Fe(NO3)3 + NxOy + (23x-9y)H2O