câu 2: Fe + 2NaCl \(\rightarrow\)FeCl2 + 2Na

3NaCl + Fe \(\rightarrow\) FeCl3 + 3Na

Cu+2H2SO4→2H2O+SO2+CuSO4

a) \(2NaCl+2H_2O-dpcmn-->2NaOH+Cl_2+H_2\)

\(Cl_2+H_2-t^0->2HCl\)

\(Fe_2O_3+3H_2-t^0->2Fe+3H_2O\)

\(Fe+2HCl-->FeCl_2+H_2\uparrow\)

\(2FeCl_2+Cl_2-->2FeCl_3\)

\(FeCl_3+3NaOH-->Fe\left(OH\right)_3\downarrow+3NaCl\)

b) \(KK-ccpd-->O_2\uparrow\)

\(4FeS_2+11O_2-t^0->2Fe_2O_3+8SO_2\uparrow\)

\(2NaCl+2H_2O-dpcmn-->2NaOH+Cl_2\uparrow+H_2\uparrow\)

\(2SO_2+O_2-t^0,V2O5-->2SO_3\)

\(SO_3+H_2O-->H_2SO_4\)

\(Fe_2O_3+3H_2SO_4-->Fe_2\left(SO_4\right)_3+3H_2O\)

\(6NaOH+Fe_2\left(SO_4\right)_3-->2Fe\left(OH\right)_3+3Na_2SO_4\)

c) \(Cu+2H_2SO_4\left(đ\right)-t^0->CuSO_4+SO_2\uparrow+2H_2O\)

\(2NaCl+H_2SO_4\left(đ\right)-->Na_2SO_4+2HCl\)

\(Fe+2HCl-->FeCl_2+H_2\uparrow\)

\(MnO_2+4HCl\left(đ\right)-->MnCl_2+2H_2O+Cl_2\)

\(2FeCl_2+Cl_2-t^0->2FeCl_3\)

\(4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

\(MnO_2+4HBr\rightarrow MnBr_2+Br_2+2H_2O\)

\(4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\)

\(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+2H_2O\)

Bài 1 : 

$a) Fe + 2HCl \to FeCl_2 + H_2$
$b) Cu(OH)_2 + 2HCl \to CuCl_2 + 2H_2O$

$c)  Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$
$d) CuCl_2 + Fe \to FeCl_2 + Cu$

Bài 2 : 

$CaO  + H_2O \to Ca(OH)_2$
$Na_2CO_3 + Ca(OH)_2 \to CaCO_3 + 2NaOH$

\(2NaCl+2H_2O-^{đpcmn}\rightarrow2NaOH+H_2+Cl_2\)

\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)

\(Cu\left(OH\right)_2-^{^{ }t^o}\rightarrow CuO+H_2O\)

a/

2NaCl\(\xrightarrow[]{đpnc}\) 2Na+ Cl₂

2Na+ 2H₂O\(\rightarrow\) 2NaOH+ H₂\(\uparrow\)

2NaOH+ CuSO₄\(\rightarrow\)Cu(OH)₂\(\downarrow\)+ Na₂SO₄

b/

Cu(OH)₂\(\xrightarrow[]{to}\) CuO+ H₂O

Bài 2:

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(Ca\left(OH\right)_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaOH\)

Bài 3

a) 

\(Na_2O+H_2O\rightarrow2NaOH\)

b)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

c) 

\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

\(Na_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+2NaOH\)

d) 

\(2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

e)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

MnO2+4HCl->MnCl2+Cl2+H2O

Fe+2HCl->FeCl2+H2

2Fe+3CL2-to>2Fecl3

b)

2NaCl+2H2O-đp\comn>2NaOH+cl2+H2

H2+Cl2-as->2HCl

Cl2+2NaOH->NaCl+NaClO+H2O

a, Fe + 2HCl ---> FeCl₂ + H₂

MnO₂ + 4HCl ---> MnCl₂ + Cl₂ + 2H₂O

2Fe + 3Cl₂ ---t^o---> 2FeCl₃

b, 2NaCl ---đpnc---> 2Na + Cl₂

2H₂O ---đp---> 2H₂ + O₂

H₂ + Cl₂ ---as---> 2HCl

2Na + 2H₂O ---> 2NaOH + H₂

2NaOH + Cl₂ ---đpkcmn---> NaCl + NaClO + H₂O

Những chất mình in đậm là nữngc chất cần điều chế nha

a)4FeS₂ + 11O₂ 2Fe₂O₃ + 8SO₂

NaCl + H₂O \(-\)^{điện phân có màng ngăn}\(\rightarrow\) NaOH + H₂ +Cl₂

\(Fe+\dfrac{3}{2}Cl_2\rightarrow FeCl_3\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)

b)Cách 1)_\(NaCl-^{đpnc}\rightarrow Na+Cl_2\)

Cách 2)_MnO₂ + 2NaCl + 2H₂SO₄ MnSO₄ + Cl₂ + Na₂SO₄ + 2H₂O

Cách 3)_NaCl + H₂O \(-\)^{điện phân có màng ngăn}\(\rightarrow\) NaOH + H₂ +Cl₂

Cách 4)_NaCl(tinh thể ) + H2SO4(đặc , nóng)----->NaHSO4 + HCl ( bay hơi lên )

Hòa tan khí HCl vào nước thu được axit clohiđric (HCl)

4HCl_đặc+ MnO₂ → Cl₂ + MnCl₂ + 2H₂O

2NaCl+2H₂O\(-\)^{điện phân có màng ngăn}\(\rightarrow\)2NaOH+H₂+Cl₂
Fe+\(\dfrac{3}{2}\)Cl₂\(\rightarrow\)FeCl₃
2FeCl₃+Fe\(\rightarrow\)3FeCl₂

\(Cl_2+H_2O\underrightarrow{\leftarrow}HCl+HClO\)

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