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

NO của chất đâu ??

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

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

3Cu+ 8HNO3-->  3Cu(NO3)2+ 2NO+ 4H2O
Cu+ 4HNO3-->Cu(NO3)2+2NO2+ 2H2O
2Fe+ 6H2SO4-->Fe2(SO4)3+ 3SO2+ 6H20
2Al+ 6H2SO4--> Al2(SO4)3+ 3SO2+ 6H2O
6Al+ 12H2SO4--> 3Al2(SO4)3+3S+ 12H2O
3Al+ 12HNO3--> 3Al(NO3)3+ 3NO+ 6H2O
10Al+ 32HNO3--> 10Al(NO3)3+ N2+ 16H2O
8Al+ 30HNO3--> 8Al(NO3)3+ 3NH4NO3+ 9H2O
 

sai r`

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

1,Mg + 2H2SO4 = MgSO4 + SO2 + 2H2O

2,Ca + 2H2SO4 = CaSO4 + SO2 + 2H2O

3 Al + 6HNO3 = Al(NO3)3 + 3NO2 + 3H2O

4,Al + 4HNO3 = Al(NO3)3 + NO + 2H2O

5,8Al + 30HNO3 = 8Al(NO3)3 + 3N2O + 15H2O

6,10Al + 36HNO3 = 10Al(NO3)3 + 3N2 + 18H2O

7,8Al + 10HNO3 = 8Al(NO3) + NH4NO3 + 3H2O

8,8Al + 15H2SO4 = 4Al2(SO4)3 + 3H2S + 12H2O

9, pthh ghi sai thiếu H

10,2Fe + 4H2SO4 = Fe2(SO4)3 + S + 4H2O

11,Fe + 6HNO3 = Fe(NO3)3 + 3NO2 + 3H2O

12,Fe + 4HNO3 = Fe(NO3)3 + NO + 2H2O

13,3Ca + 8HNO3 = 3Ca(NO3)2 + 2NO + 4H2O

14,KClO3 + 6HCl = KCl + 3Cl2 + 3H2O

15, cthh ghi sai hay sao ý

16,MnO2 + 4HCl = MnCl2 + Cl2 + 2H2O

17,Fe3O4 + 8HCl = FeCl2 + 2FeCl3 + 4H2O

18,Fe3O4 + 4H2SO4 = FeSO4 + Fe2(SO4)3 + 4H2O

19,Fe3O4 + 4CO = 3Fe + 4CO2

20, cthh ghi sai

cậu chép đề có vài chỗ ghi sai kí hiệu hóa học nữa đó, làm mik tìm mỏi cả mắt

2. 10Al + 36HNO3 -> 10Al(NO3)3 + 3N2 + 18H2O

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

4A +18HNO3 --> 4A(NO3)3 +3NO +3NO2 +9H2O (1)

vì sau phản ứng khối lượng trong bình giảm => mhh=1,42(g)

nhh=0,045(mol)

=>Mhh=31,56(g/mol)

giả sử trong 1 mol hh có x mol NO

y mol NO2

=>\(\left\{{}\begin{matrix}x+y=1\\30x+46y=31,56\end{matrix}\right.=>\left\{{}\begin{matrix}x=0,9025\left(mol\right)\\y=0,0975\left(mol\right)\end{matrix}\right.\)

lập tỉ lệ :

\(\dfrac{0,9025}{3}>\dfrac{0,0975}{3}\)

=> NO2 hết ,NO dư => tính theo NO2

theo (1) : nA=4/3nNO2=0,13(mol)

=> 5,2/M_A=0,13=> M_A=40(g/mol)

=>A:Ca

1/

1. 5Al+24HNO3->5Al(NO3)3+6NO+3NO2+12H2O

2. 3Al+48HNO3->3Al(NO3)3+3NO+3N2+24H2O