Đáp án C.

Ta có: tỉ lệ số mol giữa N₂O và N₂ là 3: 2

Pt sau cân bằng: 44Al + 162HNO₃ ®  44Al(NO₃)₃  + 6N₂ + 9N₂O + 81H₂O.

8Al + 30HNO₃ → 8Al(NO₃)₃ + 3N₂O + 15H₂O

8 × || Al → Al³⁺ + 3e

3 × || 2N⁺⁵ + 8e → 2N⁺⁴ (N₂O)

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

Sự oxi hóa: Al → Al⁺³ + 3e

Sự khử: 2N⁺⁵ + 10e → N₂

4Mg + 10HNO₃ → 4Mg(NO₃)₂ + N₂O + 5H₂O

4 × || Mg → Mg²⁺ + 2e

1 × || 2N⁺⁵ + 8e → 2N⁺⁴ (N₂O)

8Al + 30HNO₃   ---> 8Al(NO₃)₃ + 3N₂O + 15H₂O

10Al + 36HNO₃   ---> 10Al(NO₃)₃ + 3N₂ + 18H₂O

4Mg + 10HNO₃  ----> 4Mg(NO₃)₂ + N₂O + 5H₂O

\(8Al+30HNO_3\rightarrow8Al\left(NO_3\right)_3+3N_2O+15H_2O\\ 10Al+36HNO_3\rightarrow10Al\left(NO_3\right)_3+3N_2+18H_2O\\ 4Mg+10HNO_3\rightarrow4Mg\left(NO_3\right)_2+N_2O+5H_2O\)

 28Al + 102HNO₃ → 28Al(NO₃)₃ + 6N₂ + 3N₂O + 51H₂O

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.Al\(\rightarrow\)Al+3 +3e_______________.(5x-2y)

xN⁺⁵ +(5x-2y)e\(\rightarrow\)xN+\(\frac{2y}{x}\)______.3

\(\rightarrow\)(5x-2y)Al+(18x-6y)HNO3\(\rightarrow\)(5x-2y)Al(NO3)3+3NxOy+(9x-3y)H2O

2.M\(\rightarrow\)M⁺ⁿ +ne____.2

S⁺⁶ +2e\(\rightarrow\)S⁺⁴_____.n

\(\rightarrow\)2M+3nH2SO4\(\rightarrow\)2M(SO4)n+nSO2+3nH2O

3.M\(\rightarrow\)M+n +ne__________.(5x-2y)

xN+5 +(5x-2y)e\(\rightarrow\)xN+\(\frac{2y}{x}\) ____.n

\(\rightarrow\)(5x-2y)M+(6nx-2ny)HNO3\(\rightarrow\)(5x-2y)M(NO3)n+nNxOy+(3nx-ny)H2O

N⁺⁵ +3e\(\rightarrow\)N⁺² __________.1

2N⁺⁵ +8e\(\rightarrow\)2N⁺¹_________ .2

\(\rightarrow\)5N⁺⁵ +19e\(\rightarrow\)N⁺² +4N⁺¹______ .2

Zn\(\rightarrow\)Zn⁺² +2e ____________ .19

\(\rightarrow\)19Zn+48HNO3\(\rightarrow\)19Zn(NO3)2+2NO+4N2O+24H2O

2N⁺⁵ +10e\(\rightarrow\)2N0 .2

2N⁺⁵+ 8e\(\rightarrow\)2N⁺¹ .1

\(\rightarrow\)6N+5 +28e\(\rightarrow\)4N0 +2N+1 .3

Al\(\rightarrow\)Al+3 +3e _________.28

\(\rightarrow\)28Al+102HNO3\(\rightarrow\)28Al(NO3)3+6N2+3N2O+51H2O

2Fe⁺² \(\rightarrow\)2Fe⁺³ +2e .5

Mn⁺⁷ +5e\(\rightarrow\)Mn⁺² .2

\(\rightarrow\)10FeSO4+2KMnO4+2KHSO4\(\rightarrow\)5Fe2(SO4)3+2MnSO4+2K2SO4+H2O

4. (5x-2y)AL + (18x-6y)HNO3 -------> (5x-2y)AL(NO3)3 + 3NxOy +(9x-3y)H2O

5. 2M + 2nH2SO4--------> M2(SO4)n + nSO2 + 2nH2O

6. (5x-2y)M + (6nx-2ny)HNO3 -------->(5x-2y) M(NO3)n +n NxOy + (3nx-ny)H2O

7. 11Zn + 28HNO3 -------> 11Zn( NO3)2 + 2NO + 2N2O + 14H2O

8. 46AL + 168HNO3 -------;> 46AL( NO3)3 + 9N2 + 6N2O + 84H2O

9. 10FeSO4 + 2KMnO4 + 16KHSO4 ---> 5Fe2(SO4)3 +9 K2SO4 + 2MnSO4 + 8H2O.

Cân bằng các phương trình hóa học theo các sơ đồ phản ứng sau:

a)10 Al + 36HNO₃  ---> 10Al(NO₃) + 3N₂ + 18H₂O

b) 2AL + 2Ba(OH)₂ + H_2O  ---> Ba(AlO₂)₂ + 3H₂

c) 3Fe(OH)₂ + 10HNO₃ ---> 3Fe(NO₃)₃ + NO + 8H₂O

d) MnO₂ + 2FeSO₄ + 2H₂SO₄  ---> Fe₂(SO₄)₃ + MnSO₄ + 2H₂O

đề là cân bằng pt hay j v

\(38Al+138HNO_3\rightarrow38Al\left(NO_3\right)_3+9N_2+3N_2O+69H_2O\)