a)

$2NaCl + 2H_2O \xrightarrow{dpdd,cmn} 2NaOH + H_2 + Cl_2$

b)

$Cl_2 + H_2 \xrightarrow{as} 2HCl$
$Mg + 2HCl \to MgCl_2 + H_2$
$MgCl_2 + 2NaOH \to Mg(OH)_2 + 2NaCl$

c)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$AlCl_3 + 3NaOH \to Al(OH)_3 + 3NaCl$

d)

$2H_2O \xrightarrow{đp} 2H_2 + O_2$

$4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2$
$Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O$
$FeCl_3 + 3NaOH \to Fe(OH)_3 + 3NaCl$

1/ Fe₂O₃+3CO\(\rightarrow\)2Fe+3CO₂

2/ 3AgNO₃+Al=Al(NO₃)₃+3Ag

3/ 2HCl+CaCO₃=CaCl₂+H₂O+CO₂

4/ 2C₄H₁₀+13O₂=8CO₂+10H₂O

5/6NaOH+Fe₂(SO₄)₃=2Fe(OH)₃+3Na₂SO₄

6/4FeS₂+11O₂=2Fe₂O₃+8SO₂

7/6KOH+Al₂(SO₄)₃=3K₂SO₄+2Al(OH)3

8/2CH₄+O₂+2H₂O=2CO₂+6H₂

9/8Al+3Fe₃O₄=4Al₂O₃+9Fe

10/Fe_xO_y+(x-y)CO=xFeO+(x-y)CO₂

Viết phương trình hóa học biểu diễn sơ đồ biến hóa sau:
KClO3 -->(A) -->(B) -->(C) -->(D) -->CaCO3
(Trong đó (A), (B), (C), (D) là các chất riêng biệt).
2) Cho các chất rắn sau ở dạng bột: BaO, MgO, P2O5, Na2O, CuO, Fe2O3. Nêu cách làm để nhận ra mỗi chất. Viết phương trình hóa học xảy ra (nếu có).

a)

4Al + 3O2\(\rightarrow\) 2Al2O3

Al2O3 +6HNO3\(\rightarrow\) 2Al(NO3)3 + 3H2O

Al(NO3)3 + 3NaOH \(\rightarrow\)Al(OH)3 + 3NaNO3

2Al(OH)3 \(\underrightarrow{^{to}}\) Al2O3 + 3H2O

Al2O3 +6HCl \(\rightarrow\) 2AlCl3 + 3H2O

AlCl3 không điện phân nóng chảy được \(\rightarrow\) Al2O3 \(\underrightarrow{^{\text{đpnc}}}\)2Al + \(\frac{3}{2}\)O2

b) 2Fe + 3Cl2 \(\rightarrow\) 2FeCl3

FeCl3 + 3NaOH \(\rightarrow\)Fe(OH)3 + 3NaCl

2Fe(OH)3\(\rightarrow\)Fe2O3 + 3H2O

Fe2O3 + 3CO \(\rightarrow\)2Fe + 3CO2

Fe + 2HCl\(\rightarrow\) FeCl2 + 2

FeCl2 + 2AgNO3\(\rightarrow\)Fe(NO3)2 + 2AgCl (thường ra Fe⁺³ )

c) 2Mg + O2\(\rightarrow\)2MgO

MgO + 2HCl \(\rightarrow\) MgCl2 + H2O

MgCl2 + 2NaOH \(\rightarrow\) Mg(OH)2 + 2NaCl

Mg(OH)2 + H2SO4 \(\rightarrow\)MgSO4 + 2H2O

MgSO4 + BaCl2 \(\rightarrow\) MgCl2 + BaSO4

d) Cu(OH)2\(\underrightarrow{^{to}}\) CuO + H2O

CuO + H2SO4\(\rightarrow\)CuSo4 + H2O

CuSO4 + BaCl2 \(\rightarrow\) BaSO4 + CuCl2

CuCl2 + 2AgNO3 \(\rightarrow\) Cu(NO3)2 + 2AgCl

Fe + Cu(NO3)2\(\rightarrow\) Fe(NO3)2 + Cu

Cu + \(\frac{1}{2}\)O2 \(\rightarrow\) CuO

e) 4Na + O2\(\rightarrow\) 2Na2O

Na2O + H2O \(\rightarrow\) 2NaOH

2NaOH + CO2 \(\rightarrow\)Na2CO3 + H2O

Na2CO3 + H2SO4\(\rightarrow\) Na2SO4 + CO2 + H2O

Na2SO4 + BaCl2 \(\rightarrow\)BaSO4 + 2NaCl

NaCl + AgNO3\(\rightarrow\) AgCl + NaNO3

f) Fe3O4 + 4CO\(\rightarrow\) 3Fe + 4CO2

2Fe + 3Cl2 \(\rightarrow\)2FeCl3

FeCl3 + 3AgNO3 \(\rightarrow\) Fe(NO3)3 +3AgCl

Fe(NO3)3 + 3NaOH \(\rightarrow\) Fe(OH)3 + 3NaNO3

Lần sau đăng tách riêng ra nhé

Nhìn rối mắt lắm

a) Al -> Al₂O₃-> Al(NO₃)₃-> Al(OH)₃ -> Al₂O₃ -> AlCl₃ -> Al

(1) 2A1 + O₂ Al₂O₃

(2) Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

(3) Al(NO₃)₃ + 3NaOH (vừa đủ) → 3NaNO₃ + Al (OH)₃

(4) 2Al(OH)₃ Al₂O₃+ ЗН₂О

(5) 2Al₂O₃ 4Al + 3O₂

(6) 2Al + 3Cl₂ 2AlCl₃

b) Fe -> FeCl₃ -> Fe(OH)₃ -> Fe₂O₃-> Fe -> FeCl₂ -> Fe(NO₃)₃

(1) 2Fe + 3Cl₂ → 2FeCl₃

(2) FeCl₃ + 3NaOH → 3NaCl + Fe(OH)₃

(3) 2Fe(OH)₃ → Fe₂O₃ + 3H₂O

(4) Fe₂O₃ + H₂→ Fe + H₂O

(5) 3Fe + 2HCl → 2FeCl₂ +H_2↑

(6) Fe(NO₃)₃+3NaOH ➞Fe(OH)₃ +3Na(NO₃)

dài wá mình trả lời những câu còn lại sau : thông cảm cho mk nha❗❕

1. Fe2O3 + 3CO → 2Fe + 3CO2

2. 3AgNO3 + Al → Al(NO3)3 + 3Ag

3. 2HCl + CaCO3 → CaCl2 + H2O + CO2

4. 2C4H10 + 13O2 → 8CO2 + 10H2O

5. 6NaOH + Fe2(SO4)3 → 2Fe(OH) + 3Na2SO4

6. 4FeS2 + 11O2 → 2Fe2O3 + 8SO2

7. 6KOH + Al2(SO4)3 → 3K2SO4 + 2Al(OH)3

8. CH4 + 2O2 → 2H2O + CO2

9. 8Al + 3Fe3O4 → 4Al2O3 +9Fe

10. FexOy + (y-x)CO → xFeO + (y-x)CO2

1. Fe₂O₃ + CO \(\rightarrow\) 2Fe + 3CO₂

2. 3AgNO₃ + Al \(\rightarrow\) Al(NO₃)₃ + 3Ag

3. 2HCl + CaCO₃ \(\rightarrow\) CaCl₂ + H₂O + CO₂

4. 2C₄H₁₀ + 13O₂ \(\rightarrow\) 8CO₂ + 10H₂O

5. 6NaOH + Fe₂(SO₄)₃ \(\rightarrow\) 2Fe(OH) + 3Na₂SO₄

6. 4FeS₂ + 11O₂ \(\rightarrow\) 2Fe₂O₃ + 8SO₂

7. 6KOH + Al₂(SO₄)₃ \(\rightarrow\) 3K₂SO₄ + 2Al(OH)₃

8. 2CH₄ + O₂ + 2H₂O \(\rightarrow\) 2CO₂ +6H₂

9. 8Al + 3Fe₃O₄ \(\rightarrow\) 4Al₂O₃ +9Fe

10. Fe_xO_y + (y-x)CO \(\rightarrow\) xFeO + (y-x)CO₂

chúc bn học tốt

2fe(oh)3 -to> fe2o3 + 3h2o

=>phản ứng phân huỷ

fe+2hcl -> fecl2+ h2o

ba + 2h2o -> ba(oh)2 + h2

=>phản ứng thế

 2s + 3o2 -> 2so3

a) 2Al + 3Cl2 ---> 2AlCl3
b) Fe2O3 + 3H2 ---> 2Fe + 3H2O
c) 4P + 5O2 ---. 2P2O5
d) 2Al(OH)3 ---> Al2O3 + 3H2O
e) (NH4)2CO3 + 2NaOH ---> 2Na2CO3 + NH3 + 2H2O
f) Fe + HCl ---> FeCl2 + O2 (đề sai)
g) 2KClO3 ----> 2KCl + 3O2
h) 3NaOH + FeCl3 ----> Fe(OH)3 + 3NaCl
i) ZnO + 2HCl ----> ZnCl2 + H2O
k) 2K + 2H2O ----> 2KOH + H2
m) 2Na + 2H2O ----> 2NaOH + H2
n) BaCl2 + H2SO4 -----> BaSO4 + 2HCl
o) Na2O + H2O ---> 2NaOH

a) 2Al + 3Cl₂ → 2AlCl₃

b) 2Fe₂O₃ + 6H₂ → 4Fe + 6H₂O

c)4P +5O₂ → 2P₂O₅

d) 2Al(OH)₃ → Al₂O₃ + 3H₂O

e) \(\left(NH_4\right)_2CO_3+2NaOH\rightarrow Na_2CO_3+2NH_3+2H_2O\)

f) Fe + 2HCl → FeCl₂ + H₂

g) 2KClO₃ → 2KCl + 3O₂

h) 3NaOH + FeCl₃ → Fe(OH)₃ + 3NaCl

i) ZnO + 2HCl → ZnCl₂ + H₂O

k) 2K + 2H₂O → 2KOH + H₂

m) 2Na + 2H₂O → 2NaOH + H₂

n) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

o) Na₂O + H₂O → 2NaOH

2Al+3Cl2--->2AlCl3

Fe2O3+3H2--->2Fe+3H2O

4P+5O2--->2P2O5

2Al(OH)3--->Al2O3+3H2O

(NH4)2CO3+2NaOH--->Na2CO3+2NH3+2H2O

Fe+2HCl--->FeCl2+H2

2KClO3--->2KCl+3O2

3NaOH+FeCl3-->Fe(OH)3+3NaCl

ZnO++2HCl-->ZnCl2+H2O

2K+2H2O--->2KOH+H2

2Na+2H2O--->2NaOH+H2

BaCl2+H2SO4--->BaSO4+2HCl

Na2O+H2O--->2NaOH

\(1.\)     \(Fe_2O_3+3CO\rightarrow2Fe+3CO_2\)

\(2.\)     \(3AgNO_3+Al\rightarrow Al\left(NO_3\right)_3+3Ag\)

\(3.\)     \(2HCl+CaCO_3\rightarrow CaCl_2+H_2O+CO_2\)

\(4.\)     \(2C_4H_{10}+13O_2\rightarrow8CO_2+10H_2O\)

\(5.\)     \(6NaOH+Fe_2\left(SO_4\right)_3\rightarrow2Fe\left(OH\right)_3+3Na_2SO_4\)

\(6.\)     \(4FeS_2+11O_2\rightarrow2Fe_2O_3+8SO_2\)

\(7.\)     \(6KOH+Al_2\left(SO_4\right)_3\rightarrow3K_2SO_4+2Al\left(OH\right)_3\)

\(8.\)     \(2CH_4+O_2+2H_2O\rightarrow2CO_2+6H_2\)

\(9.\)     \(8Al+3Fe_3O_4\rightarrow4Al_2O_3+9Fe\)

\(10.\)   \(Fe_xO_y+\left(y-x\right)CO\rightarrow xFeO+\left(y-x\right)CO_2\)

a, 2Al+3H₂SO₄\(\rightarrow\)Al₂SO₄+3H₂

c. 2Fe+3Cl₂\(\rightarrow\)2FeCl₃

b. 2NaOH+ZnCl₂\(\rightarrow\).Zn(OH)₂+2NaCl

d. 3Fe+2O₂\(\rightarrow\)Fe₃O₄

\(\cdot2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\)

\(\cdot2NaOH+ZnCl_2\xrightarrow[]{}Zn\left(OH\right)_2+2NaCl\)

\(\cdot2Fe+3CL_2\xrightarrow[]{}2FeCl_3\)

\(\cdot3Fe+2O_2\xrightarrow[]{}Fe_3O_4\)