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Fe->fecl3->fe(OH)3->fe2o3->co2

Câu 1:

\(a,Al_2O_3+6HNO_3\to 2Al(NO_3)_3+3H_2O\\ b,Ca(OH)_2+Na_2CO_3\to CaCO_3\downarrow+2NaOH\\ c,Fe_3O_4+4CO\xrightarrow{t^o}3Fe+4CO_2\\ d,N_2+3H_2\xrightarrow{t^o,p,xt}2NH_3\\ e,Zn+2HCl\to ZnCl_2+H_2\\ f,2H_2+O_2\xrightarrow{t^o}2H_2O\\ g,4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ h,2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O\\ i,2Al+6HCl\to 2AlCl_3+3H_2\)

\(e,\text{Tỉ lệ: }Zn:H_2=1:1;HCl:ZnCl_2=2:1\\ f,\text{Tỉ lệ: }Al:H_2=2:3;HCl:AlCl_3=3:1\)

4K+O₂->2K₂O

2Fe(OH)₃->Fe₂O₃+3H₂O

2SO₂+O₂->2SO₃

Cu+2AgNO₃->Cu(NO₃)₂+2Ag

a. 4K+O₂ → 2K₂O
b. 2Fe(OH)₃ → Fe₂O₃ + 3H₂O
c. 2SO₂+O₂ → 2SO₃
d. Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

1) 4K + O₂ → 2K₂O (4:1:2)

2) 2Al +3S→ Al₂S₃ (2:3:1)

3) 2SO₂ + O₂ → 2 SO₃ (2:1:2)

4)Fe₂O₃ + 6HNO₃ → 2Fe(NO₃)₃ + 3H₂O (1:6:2:3)

5)2K + 2H₂O → 2KOH + H₂ (2:2:2:1)

6) Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄ (1:6:2:3)

7) C₂H₆O + 3O₂ → 2CO₂ + 3H₂O (1:3:2:3)

8) 2FeCl₂ +3Cl₂ → 2FeCl₃  (2:3:2)

9) 4NO₂ + O₂ + 2H₂O → 4HNO₃ (4:1:2:4)

10) C_nH_2n + \(\dfrac{3n}{2}\)O₂ → nCO₂ + nH₂O (1:\(\dfrac{3n}{2}\):n:n)

1. 4Na + O2 --- > 2Na2O

2. 2 Al + 3S --- > Al2S3

3. 2Fe(OH)3 + 6HCl --- > 2FeCl3 + 6H2O

4. Fe2O3 + 3H2SO4 --- > Fe2(SO4)3 + 3H2O

5 . 2NaOH + Cu(NO3)2 --- > 2NaNO3 + Cu(OH)2

6 . 4P + 5O2 --- > 2P2O5

1. 4Na + O2 --- > 2Na2O

2. 2Al + 3S --- > Al2S3

3. Fe(OH)3 + 3HCl --- > FeCl3 + 3H2O

4. Fe2O3 + 3H2SO4 --- > Fe2(SO4)3 + 3H2O

5 . 2NaOH + Cu(NO3)2 --- > 2NaNO3 + Cu(OH)2

6 . 4P + 5O2 --- > 2P2O5

\(1,4Na+O_2\underrightarrow{t^o}2Na_2O\\ 2,2Al+3S\rightarrow Al_2S_3\\ 3,Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

\(4,Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(5,2NaOH+Cu\left(NO_3\right)_2\rightarrow2NaNO_3+Cu\left(OH\right)_2\)

\(6,4P+5O_2\underrightarrow{t^o}2P_2O_5\)

1) \(4K+O_2\rightarrow2K_2O\)

2) \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

3) \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

4) \(2CO+O_2\rightarrow2CO_2\)

Phản ứng hoá hợp:

a)\(2Mg+O_2\underrightarrow{t^0}2MgO\)

c)\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

m)\(H_2O+Na_2O\xrightarrow[]{}2NaOH\)

o)\(K_2O+H_2O\xrightarrow[]{}2KOH\)

Phản ứng thế:

e)\(Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\)

g)\(2K+2H_2O\xrightarrow[]{}2KOH+H_2\)

h)\(HgO+H_2\underrightarrow{t^0}Hg+H_2O\)

n)\(Fe_3O_4+4H_2\underrightarrow{t^0}3Fe+4H_2O\)

Phản ứng trao đổi:

b)\(CaO+2HNO_3\xrightarrow[]{}Ca\left(NO_3\right)_2+H_2O\)

Phản ứng oxit-hoá khử

\(CH_4+2O_2\xrightarrow[]{}CO_2+2H_2O\)

$C_nH_{2n} + \dfrac{3n}{2}O_2 \xrightarrow{t^o} nCO_2 + nH_2O$
$C_nH_{2n+3}N + \dfrac{3n+1,5}{2}O_2 \xrightarrow{t^o} nCO_2 + (n + 1,5)H_2O + 0,5N_2$

$C_nH_{2n+2}O + \dfrac{3n}{2}O_2 \xrightarrow{t^o} nCO_2 + (n + 1)H_2O$

$Fe + CuSO_4 \to FeSO_4 + Cu$
$4Al(NO_3)_3 \xrightarrow{t^o} 2Al_2O_3 + 12NO_2 + 3O_2$
$3Fe_3O_4 + 8Al \xrightarrow{t^o} 9Fe + 4Al_2O_3$

$3Fe_xO_y + 2yAl \xrightarrow{t^o} yAl_2O_3 + 3xFe$

bài 1 : Lập các PTHH sau :

a. 4K + O₂ ==> 2K₂O

b. Al₂0₃ + 6HNO₃ ==> 2AI(NO₃)₃ + 3H₂O

c. 2Mg + O₂ ==> 2MgO

Bài 2 : Lập các PTHH sau :

a. FeCl₃ + 3KOH ==> Fe(OH)₃ + 3KCl

b. Mg(OH)₂ + 2HCl ==> MgCl₂ + 2H₂O

c. Al(OH)₃ + 3H₂SO₄ ==> AI(SO₄)₃ + 3H₂O

d. CuO + H₂SO₄ ==> CuSO₄ + H₂O