$\rm u)S + 2HNO_3 \rightarrow H_2SO_4 + 2NO \uparrow$

$\rm v)3Ba(NO_3)_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 \downarrow + 2Al(NO_3)_3$
$\rm w)3Ba(OH)_2 + P_2O_5 \rightarrow Ba_3(PO_4)_3 \downarrow + 3H_2O$
$\rm x)2Fe_2O_3 + 3C \xrightarrow{t^o} 4Fe + 3CO_2 \uparrow$

1) Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

2) Al + 4HNO₃ → Al(NO₃)₃ + NO + 2H₂O

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

4) 2Fe + 6H₂SO₄ → Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

`Fe + 2HCl -> FeCl_2 + H_2`

`2Na + 2HCl -> 2NaCl + H_2`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`2Al(OH)_3 + 3H_2SO_4 -> Al_2(SO_4)_3 + 6H_2O`

Fe + 2HCl → FeCl₂ + H₂
2Na + 2HCl → 2NaCl + H₂
2Al + 6HCl → 2AlCl₃ + 3H₂
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(2Na+2HCl\rightarrow2NaCl+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

$\mathrm{1)\; 3M}+\mathrm{4nH}N{O}_3\to \mathrm{3M}{\left(N{O}_3\right)}_n+\mathrm{nN}O+{\mathrm{2nH}}_{2}O$

M⁰ → M⁺ⁿ + ne |3

N⁺⁵ + 3e → N⁺² | n

Tương tự:

$\mathrm{2)\; 2M}+{\mathrm{2nH}}_{2}S{O}_4\to M_2{\left(S{O}_4\right)}_n+\mathrm{nS}{O}_2+{\mathrm{2nH}}_{2}O$

$\mathrm{3)\; 8M}+\mathrm{30H}N{O}_3\to \mathrm{8M}{\left(N{O}_3\right)}_3+{\mathrm{3N}}_{2}O+{\mathrm{15H}}_{2}O$

$\mathrm{M0\; +\; 3e}$→ M⁺³ | 8

$\mathrm{2N+5\; +\; 8e}$→ N₂⁺¹O | 3

$\mathrm{4)\; 2F}{e}_x{O}_y+\mathrm{2yH2}S{O}_4\to \mathrm{xF}{e}_2{\left(S{O}_4\right)}_{\frac{2y}{x}}+{\mathrm{2yH}}_{2}O$

$P$hương trình này không phải pt oxi hóa khử nên cứ cân bằng nhẩm bình thường

$\mathrm{các\; câu\; còn\; lại\; tương\; tự}$

a) \(K_2O+H_2O->2KOH\)

b) \(2Fe+3Cl_2->2FeCl_3\)

c) \(2Al\left(OH\right)_3+3H_2SO_4->Al_2\left(SO_4\right)_3+6H_2O\)

d) \(4FeS_2+11O_2->2Fe_2O_3+8SO_2\)

e)\(2Al+6HCl->2AlCl_3+3H_2\)

f) \(2KClO_3->KCl+3O_2\)

g) \(4P+5O_2->2P_2O_5\)

h) \(3Fe+2O_2->Fe_3O_4\)

i) \(2Al+3CuSO_4->Al_2\left(SO_4\right)_3+3Cu\)

j) \(2K+2H_2O->2KOH+H_2\)

1)

(5x-2y)FeO+(16x-6y)HNO₃\(\rightarrow\)(5x-2y)Fe(NO₃)₃+N_xO_y+(8x-3y)H₂O

2)

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

3)

3xFe₂O₃+(6x-4y)Al\(\rightarrow\)6Fe_xO_y+(3x-2y)Al₂O₃

4)

3Fe_mO_n+(12m-2n)HNO₃\(\rightarrow\)3mFe(NO₃)₃+(3m-2n)NO+(6m-n)H₂O

5)

(10-2x)FeS₂+(60-6x)HNO₃\(\rightarrow\)(10-2x)Fe(NO₃)₃+15N₂O_x+(10+x)H₂O+(20-4x)H₂SO₄

1,

2,

3,Cu+2H2SO4→2H2O+SO2+CuSO4

4,Fe+4HNO3→2H2O+NO+Fe(NO3)3

5,

6,4 Zn0 + 5 H2SO4 → 4 ZnSO4 + H2S + 4 H2O

7,4Mg + 10HNO3->4Mg(NO3)2 + NH4NO3 + 3H2O

8,2 KMnO4 + K2SO3 + 2 KOH → 2 K2MO2O4 + K2SO4 + H2O

9,2 KMnO4 + 10 FeSO4 + 8 H2SO4 → 2 Mn(SO4) + 5 Fe2(SO4)3 +K2SO4 + 8 H2O

11 câu?

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

$C_x{H}_y{O}_z$+$\left(x+\frac{y}{4}-\frac{z}{2}\right)O_2$---->$xC{O}_2+\frac{y}{2}{H}_{2}O$

3)

2Al + 6H2SO4 → Al2(SO4)3 + 3SO2 + 6H2O

Cân bằng CxHyOz + O2 ------> CO2 + H2O - Thuy Kim