`n_[FeO]=[14,4]/72=0,2(mol)`

`a)PTHH:`

`FeO + 2HCl -> FeCl_2 + H_2 O`

`0,2`       `0,4`            `0,2`                          `(mol)`

`b)m_[FeCl_2]=0,2.127=25,4(g)`

`c)m_[dd HCl]=[0,4.36,5]/10 .100=146(g)`

`d)C%_[FeCl_2]=[25,4]/[14,4+146].100~~15,84%`

\(MgO+2HCl-->MgCl_2+H_2\)

\(Fe\left(OH\right)_3+3HCl-->FeCl_3+3H_2O\)

\(NaCO_3+2HCl-->2NaCl+H_2O+CO_2\)

\(BaSO_3+2HCl-->BaCl_2+H_2O+SO_2\)

\(KHCO_3+HCl-->KCl+CO_2+H_2\)

NaNO3 khong pứ vs HCl

\(AgNO_3+HCl-->AgCl+HNO_3\)

`2Fe + 6H_2 SO_[4(đ,n)] -> Fe_2(SO_4)_3 + 3SO_2 \uparrow + 6H_2 O`

`0,05`        `0,15`                               `0,025`                                     `(mol)`

`Cu + 2H_2 SO_[4(đ,n)] -> CuSO_4 + SO_2 \uparrow + 2H_2 O`

`0,225`     `0,45`                         `0,225`                                          `(mol)`

`n_[SO_2]=[6,72]/[22,4]=0,3(mol)`

Gọi `n_[Fe]=x` ; `n_[Cu]=y`

`=>` $\begin{cases} \dfrac{3}{2}x+y=0,3\\56x+64y=17,2 \end{cases}$

`<=>` $\begin{cases}x=0,05\\y=0,225 \end{cases}$

  `@m_[Fe_2(SO_4)_3]=0,025.400=10(g)`

  `@m_[CuSO_4]=0,225.160=36(g)`

  `@m_[dd H_2 SO_4]=[(0,15+0,45).98]/80 .100=73,5(g)`

Sửa đề: 80% ---> 98% (80% chưa đặc nên không giải phóng SO₂ được)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\)

\(\rightarrow56a+64b=17,2\left(1\right)\)

PTHH: 

\(2Fe+6H_2SO_{4\left(đặc,nóng\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

a------>3a------------------->0,5a--------------->1,5a

\(Cu+2H_2SO_{4\left(đặc,nóng\right)}\rightarrow CuSO_4+SO_2\uparrow+2H_2O\)

b----->2b------------------->b------------->b

\(\rightarrow1,5a+b=\dfrac{6,72}{22,4}=0,3\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\rightarrow\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,225\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Fe_2\left(SO_4\right)_3}=0,5.0,05.400=10\left(g\right)\\m_{CuSO_4}=0,225.160=36\left(g\right)\\m_{ddH_2SO_4}=\dfrac{\left(0,05.3+0,225.2\right).98}{98\%}=60\left(g\right)\end{matrix}\right.\)

\(a,m_{Na_2CO_3}=\dfrac{500.20}{100}=100\left(g\right)\\ \rightarrow n_{Na_2CO_3}=\dfrac{100}{106}=\dfrac{50}{53}\left(mol\right)\)

PTHH: \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+CO_2\uparrow+H_2O\)

              \(\dfrac{50}{53}\)------->\(\dfrac{100}{53}\)--------------->\(\dfrac{100}{53}\)-------------->\(\dfrac{50}{53}\)

\(b,m_{axit}=\dfrac{100}{53}.60=\dfrac{6000}{53}\left(g\right)\\ c,m_{dd}=500+400-\dfrac{50}{53}.44=\dfrac{45500}{53}\left(g\right)\\ m_{CH_3COONa}=\dfrac{100}{53}.82=\dfrac{8200}{53}\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{\dfrac{8200}{23}}{\dfrac{45500}{23}}.100\%=18,02\%\)

`n_[Zn]=13/65=0,2(mol)`

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,2`   `0,4`                             `0,2`       `(mol)`

`a)V_[H_2]=0,2.22,4=4,48(l)`

`b)C_[M_[HCl]]=[0,4]/[0,2]=2(M)`

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,2        0,4                       0,2 
\(V_{H_2}=0,2.22,4=4,48\left(L\right)\\ C_{M\left(HCl\right)}=\dfrac{0,4}{0,2}=2M\)

C50 :

X1 : Ba(OH)₂ 

X2 : Ca(HCO₃)₂

\(Ba\left(OH\right)_2+Ca\left(HCO_3\right)_2\rightarrow BaCO_3+CaCO_3+2H_2O\)

X3 : Ca 

X4 : H₂O

\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

X5 : FeCl₃

X6 : Na₂CO₃

\(2FeCl_3+3Na_2CO_3+3H_2O\rightarrow2Fe\left(OH\right)_3+3CO_2+6NaCl\)

51 :

\(\left(1\right)2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

\(\left(2\right)FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\)

\(\left(3\right)Fe\left(NO_3\right)_3+NaOH\rightarrow Fe\left(OH\right)_3+NaNO_3\)

\(\left(4\right)2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

\(\left(5\right)Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(\left(6\right)Fe+HNO_{3\left(đ,n\right)}\rightarrow Fe\left(NO_3\right)_3+NO_2+H_2O\)

\(\left(7\right)2FeCl_3+3H_2SO_{4\left(đ,n\right)}\rightarrow Fe_2\left(SO_4\right)_3+6HCl\)

\(\left(8\right)2Fe\left(NO_3\right)_3+Zn\rightarrow2Fe\left(NO_3\right)_2+Zn\left(NO_3\right)_2\)

\(\left(9\right)Fe\left(OH\right)_3+3HNO_3\rightarrow Fe\left(NO_3\right)_3+H_2O\)

\(\left(10\right)Fe_2O_3+HNO_3\rightarrow Fe\left(NO_3\right)_3+H_2O\)

Câu 50:

a) X₁, X₂ là Ca(HCO₃)₂, Ba(OH)₂

Ca(HCO₃)₂ + Ba(OH)₂ ---> CaCO₃↓ + BaCO₃↓ + 2H₂O

b) X₃, X₄ là Ca, H₂O

Ca + 2H₂O ---> Ca(OH)₂ + H₂

c) X₅, X₆ là FeCl₃, Na₂CO₃

2FeCl₃ + 3Na₂CO₃ + 3H₂O ---> 2Fe(OH)₃↓ + 6NaCl + 3CO₂↑

___________________________________________________

51)

(1) 2Fe + 3Cl₂ --t^o--> 2FeCl₃

(2) FeCl₃ + 3AgNO₃ ---> 3AgCl↓ + Fe(NO₃)₃ 

(3) Fe(NO₃)₃ + 3NaOH ---> Fe(OH)₃↓ + 3NaNO₃

(4) 2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

(5) Fe₂O₃ + 3CO --t^o--> 2Fe + 3CO₂↑

(6) Fe + 3AgNO₃ ---> Fe(NO₃)₃ + 3Ag↓

(7) 2FeCl₃ + 3H₂SO_{4(đặc, nóng)} ---> Fe₂(SO₄)₃ + 6HCl

(8) Fe + 2Fe(NO₃)₃ ---> 3Fe(NO₃)₂

(9) Fe(OH)₃ + 3HNO₃ ---> Fe(NO₃)₃ + 3H₂O

(10) Fe₂O₃ + 6HNO₃ ---> 2Fe(NO₃)₃ + 3H₂O

___________________________________________________

52)

a) 

2Na + 2HCl ---> 2NaCl + H₂

\(2NaCl+2H_2O\xrightarrow[\text{có màng ngăn}]{\text{điện phân}}2NaOH+Cl_2\uparrow+H_2\uparrow\)

NaOH + HNO₃ ---> NaNO₃ + H₂O

Cu + 2NaNO₃ + 2H₂SO_{4 (đặc, nóng)} ---> Na₂SO₄ + CuSO₄ + 2NO₂↑ + 2H₂O

b) 

4Na + O₂ --t^o--> 2Na₂O

Na₂O + H₂O ---> 2NaOH

2NaOH + CO₂ ---> Na₂CO₃ + H₂O

Na₂CO₃ + CO₂ + H₂O ---> 2NaHCO₃

NaHCO₃ + NaOH ---> Na₂CO₃ + H₂O

Na₂CO₃ + 2HCl ---> 2NaCl + CO₂↑ + H₂O

NaCl + AgNO₃ ---> AgCl↓ + NaNO₃

c)

4FeS₂ + 11O₂ --t^o--> 2Fe₂O₃ + 8SO₂↑

\(2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3\)

SO₃ + H₂O ---> H₂SO₄

2H₂SO_{4(đặc, nóng)} + Zn ---> ZnSO₄ + SO₂↑ + 2H₂O

SO₂ + Br₂ + 2H₂O ---> H₂SO₄ + 2HBr

H₂SO₄ + Ba(OH)₂ ---> BaSO₄↓ + 2H₂O

d)

4Al + 3O₂ --t^o--> 2Al₂O₃

\(2Al_2O_3\xrightarrow[criolit\left(Na_3AlF_6\right)]{\text{điện phân nóng chảy}}4Al+3O_2\uparrow\)

2NaOH + 2Al + 2H₂O ---> 2NaAlO₂ + 3H₂↑

NaAlO₂ + HCl + H₂O ---> Al(OH)₃↓ + NaCl

2Al(OH)₃ --t^o--> Al₂O₃ + 3H₂O

Al₂O₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂O

Al₂(SO₄)₃ + 3BaCl₂ ---> 3BaSO₄↓ + 2AlCl₃ 

\(2AlCl_3\xrightarrow[]{\text{điện phân dung dịch}}2Al+3Cl_2\uparrow\)

e)

Zn + 2NaOH ---> Na₂ZnO₂ + H₂↑

2Zn + O₂ --t^o--> 2ZnO

ZnO + CO --t^o--> Zn + CO₂

ZnO + 2NaOH ---> Na₂ZnO₂ + H₂O 

Na₂ZnO₂ + 4HCl_(dư) ---> 2NaCl + ZnCl₂ + 2H₂O

ZnCl₂ + 4NaOH_(dư) ---> Na₂ZnO₂ + 2NaCl + 2H₂O

ZnCl₂ + 2NaOH ---> Zn(OH)₂↓ + 2NaCl

Zn(OH)₂ --t^o--> ZnO + H₂O

g)

\(N_2+O_2\xrightarrow[]{3000^oC}2NO\uparrow\)

2NO + O₂ ---> 2NO₂↑

4NO₂ + O₂ + 2H₂O ---> 4HNO₃ 

2HNO₃ + CuO ---> Cu(NO₃)₂ + H₂O

3Cu + Cu(NO₃)₂ + 8HCl ---> 4CuCl₂ + 2NO↑ + 4H₂O

CuCl₂ + Ca(OH)₂ ---> Cu(OH)₂↓ + CaCl₂

Cu(OH)₂ --t^o--> CuO + H₂O

CuO + H₂ --t^o--> Cu + H₂O

Cu + 2FeCl₃ ---> 2FeCl₂ + CuCl₂ 

h) X là Al 

(1) \(2Al_2O_3\xrightarrow[criolit\left(Na_3AlF_6\right)]{\text{điện phân nóng chảy}}4Al+3O_2\uparrow\)

(2) 2Al + Ca(OH)₂ + 2H₂O ---> Ca(AlO₂)₂ + 3H₂↑

(3) Ca(AlO₂)₂ + CO₂ + H₂O ---> CaCO₃↓ + 2Al(OH)₃↓ + 3H₂O

(4) Al(OH)₃ + 3HCl ---> AlCl₃ + 3H₂O

(5) AlCl₃ + 3AgNO₃ ---> 3AgCl↓ + Al(NO₃)₃

(6) 2Al(NO₃)₃ + 3Mg ---> 3Mg(NO₃)₂ + 2Al↓

52 ) . 

a ) .

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

\(2NaCl+2H_2O\rightarrow2NaOH+H_2+Cl_2\)

\(2NaOH_{\left(loãng\right)}+2NO_2\rightarrow NaNO_3+H_2O+NaNO_2\)

\(Cu+2NaNO_3+2H_2SO_{4\left(đặc\right)}\rightarrow Na_2SO_4+CuSO_4+2NO_2+2H_2O\)

\(2NO_2+2NaOH\rightarrow NaNO_3+H_2O+NaNO_2\)

\(_{---------------------------------------------------}\)

b ) .

\(4Na+O_2\underrightarrow{t^o}2Na_2O\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

\(Na_2CO_3+H_2O\rightarrow NaHCO_3+NaOH\)

\(NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\)

\(Na_2CO_3+HCl\rightarrow NaCl+NaHCO_3\)

\(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\)

\(_{-----------------------------------------------}\)

c ) . 

\(FeS_2+2O_2\underrightarrow{t^o}Fe+2SO_2\)

\(2SO_2+O_2\underrightarrow{t^o}2SO_3\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(H_2SO_4+H_2\underrightarrow{t^o}SO_2+2H_2O\)

\(2SO_2+2H_2O+O_2\underrightarrow{t^o}2H_2SO_4\)

\(H_2SO_4+Ba\rightarrow BaSO_4+H_2\)

\(_{------------------------------------}\)

d ) . 

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

\(2Al_2O_3\underrightarrow{criolit}4Al+3O_2\)

\(Al+NaO_2\rightarrow NaAlO_2\)

\(NaAlO_2+H_2O+HCl\rightarrow Al\left(OH\right)_3+NaCl\)

\(2Al\left(OH\right)_3\underrightarrow{t^o\left(>575^oC\right)}Al_2O_3+2H_2O\)

\(Al_2O_3+2H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\)

\(2AlCl_3+3Mg\rightarrow2Al+3MgCl_2\)

\(_{-------------------------------------------------}\)

e ) . 

\(Na_2ZnO_2+H_2\leftarrow Zn+2NaOH\)

\(2Znn+O_2⇌2ZnO\)

\(ZnO+2NaOH\rightarrow Na_2ZnO_2+H_2O\)

điều chế ZnCl2 từ Na2ZnO2 cần 2 bước ý bn=)

b1 : \(Na_2ZnO_2+CO_2+H_2O\rightarrow2NaHCO_3+Zn\left(OH\right)_2\)

b2 : cho Zn ( OH)2 td vs HCl

\(Zn\left(OH\right)_2+2HCl\rightarrow ZnCl_2+2H_2O\)

\(ZnCl_2+2KOH\rightarrow2KCl+Zn\left(OH\right)_2\)

\(Zn\left(OH\right)_2\underrightarrow{t^o\left(100-250^oC\right)}ZnO+H_2O\)

\(_{---------------------------------------------}\)

câu f mất tiêu hã bà :>

g ) . 

\(N_2+O_2\underrightarrow{t^o}2NO\)

\(2NO+O_2\underrightarrow{t^o}2NO_2\)

\(3NO_2+H_2O\rightarrow2HNO_3+NO\)

\(2HNO_3+CuO\rightarrow Cu\left(NO_3\right)_2+H_2O\)

\(Cu\left(NO_3\right)_2+3Cu+8HCl\rightarrow4H_2O+2NO+4CuCl_2\)

\(CuCl_2+Ca\left(OH\right)_2\rightarrow Cu\left(OH\right)_2+CaCl_2\)

\(Cu\left(OH\right)_2\underrightarrow{t^o\left(40-80^oC\right)}CuO+H_2O\)

\(3CuO+2Al\underrightarrow{t^o\left(ánh.sáng\right)}3Cu+Al_2O_3\)

 Câu 47:

4FeS₂ + 11O₂ --t^o--> 8SO₂↑ + 2Fe₂O₃ 

SO₂ + 2H₂S ---> 3S↓ + 2H₂O

S + Fe --t^o--> FeS

FeCl₂ + 2NaOH ---> Fe(OH)₂↓ + 2NaCl

2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

Fe + 2HCl ---> FeCl₂ + H₂S↑
4Fe(OH)₂ + O₂ + 2H₂O ---> 4Fe(OH)₃↓

___________________________________________________

Câu 48:

FeS + 2HCl ---> H₂S↑ + FeCl₂ 

2H₂S + SO₂ ---> 3S↓ + 2H₂O

2FeCl₃ + 2KI ---> 2FeCl₂ + I₂↓ + 2KCl

H₂S + CuSO₄ ---> CuS↓ + H₂SO₄

2FeCl₂ + Cl₂ --t^o--> 2FeCl₃

___________________________________________________

Câu 49:

a) X₁, X₂ là KMnO₄, HCl

2KMnO₄ + 16HCl --đun nhẹ--> 5Cl₂↑ + 2MnCl₂ + 2KCl + 8H₂O

b) X₃, X₄, X₅ là H₂S, Cl₂, H₂O

H₂S + 4Cl₂ + 4H₂O ---> H₂SO₄ + 8HCl

c) A₁, A₂ là H₂S, O₂

2H₂S + 3O₂ --t^o--> 2H₂O + 2SO₂↑

d) X, Y là H₂PO₄, OH

Ca(H₂PO₄)₂ + 2Ca(OH)₂ ---> Ca₃(PO₄)₂ + 4H₂O

e) D₁, D₂, D₃ là NaCl, H₂SO₄, KMnO₄

10NaCl + 8H₂SO₄ + 2KMnO₄ ---> 5Cl₂↑ + 2MnSO₄ + 5Na₂SO₄ + K₂SO₄ + 8H₂O

f) G₁, G₂, G₃ là CaCO₃, KOH, H₂O

KHCO₃ + Ca(OH)₂ ---> CaCO₃↓ + KOH + H₂O

g) L₁, L₂, L₃ là Al₂(SO₄)₃, K₂SO₄, H₂O

Al₂O₃ + 6KHSO₄ ---> Al₂(SO₄)₃ + 3K₂SO₄ + 3H₂O

Quy đổi hh thành A₂O_n: x (mol)

=> dd A chứa A(OH)_n: 2x (mol)

1/10 dd A chứa \(A\left(OH\right)_n:0,2.x\left(mol\right)\)

\(n_{HCl}=0,06.1=0,06\left(mol\right)\)

PTHH: A(OH)_n + nHCl --> ACl_n + nH₂O

             0,2x--->0,2xn-->0,2x

=> 0,2xn = 0,06

=> xn = 0,3

\(m_{muối}=m_{ACl_n}=0,2x.\left(M_A+35,5n\right)=0,2x.M_A+7,1xn\left(g\right)\) 

= 0,2x.M_A + 2,13 (g) (1)

Có: \(n_{A_2O_n}=x=\dfrac{21,2}{2.M_A+16n}\left(mol\right)\)

=> 2x.M_A + 16nx = 21,2 

=> x.M_A = 8,2 (2)

(1)(2) => m_muối = 3,77 (g)

Bài 14:

A, B, D, E, F lần lượt là KCl, O₂, KOH, Cl₂, H₂O

\(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\\ 2KCl+2H_2O\xrightarrow[\text{có màng ngăn}]{\text{điện phân}}2KOH+Cl_2\uparrow+H_2\uparrow\\ 2KOH+Cl_2\rightarrow KCl+KClO+H_2O\)

Bài 15:

A, B, D, E, F, G, X, Y, Z lần lượt là Fe₃O₄, HCl, FeCl₂, Cl₂, FeCl₃, KOH, Fe₂O₃, H₂, CO, C

Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

Fe₃O₄ + 4CO --t^o--> 3Fe + 4CO₂

Fe₃O₄ + 4C --t^o--> 3Fe + 4CO

Fe + 2HCl ---> FeCl₂ + H₂

2FeCl₂ + Cl₂ --t^o--> 2FeCl₃

FeCl₃ + 3KOH ---> Fe(OH)₃ + 3KCl

2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

_________________________________________________

Bài 16:

a)

A, B là FeCl₂, Fe(OH)₂

(1) Fe + 2HCl ---> FeCl₂ + H₂

(2) FeCl₂ + 2LiOH ---> Fe(OH)₂ + 2LiCl

(3) Fe(OH)₂ + H₂SO₄ ---> FeSO₄ + 2H₂O

b)

C, D, E là CuCl₂, Cu(OH)₂, CuO

(1) CuCl₂ + Ba(OH)₂ ---> Cu(OH)₂ + BaCl₂

(2) Cu(OH)₂ --t^o--> CuO + H₂O

(3) CuO + CO --t^o--> Cu + CO₂

_________________________________________________

20)

A, B, C, D, E, F, G lần lượt là Fe, FeCl₂, FeCl₃, Fe(OH)₂, Fe(OH)₃, FeO, Fe₂O₃

Fe + 2HCl ---> FeCl₂ + H₂

2Fe + 3Cl₂ --t^o--> 2FeCl₃

2FeCl₃ + Fe ---> 3FeCl₂

FeCl₂ + Ca(OH)₂ ---> Fe(OH)₂ + CaCl₂

FeCl₃ + 3NaOH ---> Fe(OH)₃ + 3NaCl

4Fe(OH)₂ + 2H₂O + O₂ ---> 4Fe(OH)₃ 

\(Fe\left(OH\right)_2\xrightarrow[\text{chân không}]{t^o}FeO+H_2O\)

2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

Fe₂O₃ + CO --t^o--> 2FeO + CO₂

FeO + H₂ --t^o--> Fe + H₂O

Fe₂O₃ + 2Al --t^o--> 2Fe + Al₂O₃

_________________________________________________

A, B, C, D, E lần lượt là Cu, CuS, CuO, CuCl₂, CuSO₄

Dãy trên:

Cu + S --t^o--> CuS

2CuS + 3O₂ --t^o--> 2CuO + 2SO₂

CuO + 2HCl ---> CuCl₂ + H₂O

\(CuCl_2\xrightarrow[]{\text{điện phân dung dịch}}Cu+Cl_2\uparrow\)

Dãy dưới:

2CuS + 3O₂ --t^o--> 2CuO + 2SO₂

CuO + H₂ --t^o--> Cu + H₂O

Cu + 2H₂SO_{4(đặc, nóng)} ---> CuSO₄ + SO₂ + 2H₂O

CuSO₄ + Fe ---> FeSO₄ + Cu

Bài 10:

A, B, C, D, E lần lượt là Cl₂, HCl, MnCl₂, NaCl, NaClO

(1) H₂ + Cl₂ --as--> 2HCl

(2) 4HCl + MnO₂ ---> MnCl₂ + Cl₂↑ + 2H₂O

(3) Cl₂ + 2NaOH ---> NaCl + NaClO + H₂O

______________________________________________________

Bài 11:

a)

A, B, C, D, E lần lượt là CO, CO₂, Ca(OH)₂, CaCO₃, H₂O

3CO + Fe₂O₃ --t^o--> 2Fe + 3CO₂

2CO₂ + Ca(OH)₂ ---> Ca(HCO₃)₂

Ca(HCO₃)₂ + Ca(OH)₂ ---> 2CaCO₃↓ + 2H₂O

Ca(HCO₃)₂ --t^o--> CaCO₃ + CO₂ + H₂O

b)

A, B, C, D, E lần lượt là H₂, HCl, FeCl₂, NaOH, Fe(OH)₂

Cl₂ + H₂ --as--> 2HCl

2HCl + Fe ---> FeCl₂ + H₂↑

FeCl₂ + 2NaOH ---> Fe(OH)₂↓ + 2NaCl

Fe(OH)₂ + 2HCl ---> FeCl₂ + 2H₂O

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Bài 12:

\(C\rightarrow CO\rightarrow CO_2\rightarrow Ca\left(HCO_3\right)_2\rightarrow CaCO_3\rightarrow CaCl_2\)

2C + O₂ --t^o--> 2CO

2CO + O₂ --t^o--> 2CO₂

2CO₂ + Ca(OH)₂ ---> Ca(HCO₃)₂

Ca(HCO₃)₂ --t^o--> CaCO₃ + CO₂ + H₂O

CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O

Xin lỗi, mình không để ý thấy bài 13

Bài 13:

A, B, C, D lần lượt là NaCl, Cl₂, HCl, FeCl₂

(1) Cl₂ + 2Na --t^o--> 2NaCl

(2) \(2NaCl+2H_2O\xrightarrow[\text{có màng ngăn}]{\text{điện phân}}2NaOH+Cl_2\uparrow+H_2\uparrow\)

(3) Cl₂ + H₂ --as--> 2HCl

(4) 4HCl + MnO₂ ---> MnCl₂ + Cl₂ + 2H₂O

(5) 2HCl + Fe ---> FeCl₂ + H₂

(6) FeCl₂ + 2NaOH ---> Fe(OH)₂ + 2NaCl

(7) NaOH + HCl ---> NaCl + H₂O

(8) NaCl + H₂SO_{4(đặc, nóng)} ---> NaHSO₄ + HCl