Mới lớp 8 mà cân bằng electron rồi à bạn :3

a) 1:2:1:2:2

b) 1:2:1:1:2

c) 2:16:2:2:5:8

d) 10:2:8:5:1:2:8

e) 8:30:8:3:9

Cân bằng các PTHH sau :

a) ${\left(N{H}_4\right)}_{2}C{O}_3+2NaOH\to N{a}_{2}C{O}_3+2N{H}_3+2H_{2}O$

b)$Cu+2H_{2}S{O}_4\to CuS{o}_4+S{O}_2+2H_{2}O$

c)2 $KMn{O}_4+16HCl\to 2KCl+2MnC{l}_2+5C{l}_2+8H_{2}O$

d)10$FeS{O}_4+2KMn{O}_4+8H_{2}S{O}_4\to F{e}_2{\left(S{O}_4\right)}_3+2K_{2}S{O}_4+8MnS{O}_4+{\mathrm{H2}}^{}O$

e)8$Al+30HN{O}_3\to 8Al{\left(N{O}_3\right)}_3+3N{H}_{4}N{O}_3+9H_{2}O$

a) 4Na + O₂ → 2Na₂O

b) 2Cu + O₂ → 2CuO

c) (4x + y)O₂ + 2C_xH_y → 2xCO₂ + yH₂O

e) 8H₂SO₄ + 2KMnO₄ + 10FeSO₄ → 5Fe₂(SO₄)₃ + 8H₂O + 2MnSO₄ + K₂SO₄

f) 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

a/ 4Na + O₂ ===> 2Na₂O

b/ 2Cu + O₂ ===> 2CuO

c/ 2C_xH_y +(4x+y) O₂ ===> 2xCO₂ + yH₂O

d/ 2C_xH_yO_t +\(\frac{4x+y-2t}{2}\) O₂ ===> 2xCO₂ + yH₂O

e/ 8H₂SO₄+2KMnO₄+10FeSO₄→5Fe₂(SO4)₃+8H₂O+2MnSO₄+K2SO₄

f/ 2KMnO₄ =(nhiệt)==> K₂MnO₄ + MnO₂ + O₂

\(2FeO+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)

Câu trả lời của mình :

\(2FeO+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)

\(6FeO+10H_2SO_4\rightarrow3Fe_2\left(SO_4\right)_3+S+10H_2O\)

\(2KHCO_3+Ca\left(OH\right)_2\rightarrow K_2CO_2+CaCO_3+2H_2O\)

\(Al_2O_3+6KHSO_4\rightarrow Al_2\left(SO_4\right)_3+3K_2SO_4+3H_2O\)

\(nFe_2O_3+\left(3y-2x\right)H_2\rightarrow2Fe_xO_y+\left(3y-2x\right)H_2O\)

\(2NaHSO_4+BaCO_3\rightarrow Na_2SO_4+BaSO_4+CO_2+H_2O\)

\(6H_2SO_4+2Fe\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

1, 4

2, 2

3, 8

4,

5, 8

6, 10

a)

\(2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2(X)\\ 3O_2 + 4Al \xrightarrow{t^o} 2Al_2O_3(Y)\\ Al_2O_3 + 6HCl \to 2AlCl_3(Z) + 3H_2O\\ AlCl_3 + 3KOH \to Al(OH)_3(T) + 3KCl\\ 2Al(OH)_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O\\ \)

b)

\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2(A)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ 2H_2O + 2Na \to 2NaOH(B) + H_2\\ 2NaOH \xrightarrow{đpnc} Na + O_2 + H_2 CuO + H_2 \xrightarrow{t^o} Cu(C) + H_2O\)

A: O2 B: Al2O3 C: AlCl3 D: Al(OH)3

2KClO3-t^o-->2KCl+3O2

3O2+4Al--->2Al2O3

Al2O3+6HCl--->2AlCl3+3H2O

AlCl3+3NaOH--->Al(OH)3+3NaCl

2Al(OH)3+3H2SO4--->Al2(SO4)3+6H2O

b> 2

$KMn{O}_4\underrightarrow{t^o}{K}_{2}Mn{O}_4+Mn{O}_2+O_2$

$O_2+2H_2\underrightarrow{t^o}2H_{2}O$

$H_{2}O+S{O}_3\to H_{2}S{O}_4$

$Fe+H_{2}S{O}_4\to FeS{O}_4+H_2$

$2H_2+C{O}_2{}^{}$

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

\(2Fe_xO_y+\left(6x-2y\right)H_2SO_4\rightarrow xFe_2\left(SO_4\right)_3+\left(3x-2y\right)SO_2+\left(6x-2y\right)H_2O\)

Cái bài này vừa cân bằng e vừa cân bằng theo cách đếm. ng tử nx nha !

\(aFeSO_4+bKMnO_4+cH_2SO_4\rightarrow dFe_2\left(SO_4\right)_3+eK_2SO_4+fMnSO_4+gH_2O\)

Suy ra:

\(Fe:a=2d\)

\(S:a+c=e+f+3d\Rightarrow c=d+3=g\)

\(O:4a+4b+4c+12d+4e+4f+g\) (*)

\(K:b=2e\)

\(Mn:b=f\)

\(H:2c=2g\Rightarrow c=g\)

Chọn \(e=1\Rightarrow b=1=f\)

\(\left(\text{*}\right)\Rightarrow8d+8+4\left(d+3\right)=12d+4+8+d+3\)

\(\Rightarrow8d+8+4d+12=13d+15\)

\(\Rightarrow13d-8d-4d=8+12-15\)

\(\Rightarrow d=5\Rightarrow c=g=8\)

\(\Rightarrow a=10\)

Do đó: \(10FeSO_4+2KMnO_4+8H_2SO_4\rightarrow5Fe_2\left(SO_4\right)_3+K_2SO_4+2MnSO_4+8H_2O\)

Đó nha chị! @Azue

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Fe₃O₄ + 4H₂SO₄ → FeSO₄ + Fe₂(SO₄)₃ + 4H₂O