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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.
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.
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
a) 3Mg + 8HNO3 --> 3Mg(NO3)2 + 2NO + 4H2O
b) 2Fe + 6H2SO4 --> Fe2(SO4)3 + 3SO2 + 6H2O
c) 8Mg + 10H2SO4 --> 8MgSO4 + 2H2S + 8H2O
d) 8Al + 30HNO3 --> 8Al(NO3)3 + 3N2O + 15H2O
1, 4Mg + 10HNO3 loãng------> 4Mg(NO3)2 + N2O + 5H2O
2, 10Al + 36HNO3 loãng------> 10Al(NO3)3 + 3N2 + 18H2O
3, 4Mg+ 10HNO3 loãng--------> 4Mg(NO3)2+ NH4NO3 + 3H2O
4, 3Fe3O4+ 28HNO3 loãng--------> .9Fe(NO3)3 +NO + 14H2O
5, 8FeS + 8 H2SO4 đặc------> 3Fe2(S04)3 + SO2+8H2
6, FeS2 + 18HNO3 đặc -------> Fe(NO3)3+ 15NO2 + 2H2SO4 + 7H2O
1, 4Mg+10HNO3 ----->4Mg(NO3)2+N2O+5H20
Quá trình khử: 2NO3− + 10H+ + 8e− = N2O + 5H
Quá trình oxy hoá Mg − 2e− = Mg2+
\(38Al+138HNO_3\rightarrow38Al\left(NO_3\right)_3+9N_2+3N_2O+69H_2O\)