$a) MnO_2 + 4HCl \xrightarrow{t^o} MnCl_2 + Cl_2 + 2H_2O$
$Cl_2 + Cu \xrightarrow{t^o} CuCl_2$

$CuCl_2 + 2NaOH \to Cu(OH)_2 + 2NaCl$
$2NaCl + 2H_2O \xrightarrow{đpdd, cmn} 2NaOH + H_2 + Cl_2$
$2NaOH + Cl_2 \to NaCl + NaClO + H_2O$
$NaClO + HCl \to NaCl + HClO$

b)

$2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$
$Cl_2 + H_2 \xrightarrow{ánh\ sáng} 2HCl$
$HCl + KOH \to KCl + H_2O$

$2KCl + 2H_2O \xrightarrow{đpdd, cmn} 2KOH + H_2 + Cl_2$
$3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3$

$FeCl_3 + 3AgNO_3 \to Fe(NO_3)_3 + 3AgCl$

a, Cl2 + 2Na -> (t°) 2NaCl

2NaCl + H2SO4 -> Na2SO4 + 2HCl

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

Ca + Cl2 -> (t°) CaCl2

b, 4HCl + MnO2 -> MnCl2 + 2H2O + Cl2

2Fe + 3Cl2 -> (t°) 2FeCl3

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

2NaCl + H2SO4 -> Na2SO4 + 2HCl

c, MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2

3Cl2 + 6KOH (đặc nóng) -> 5KCl + KClO3 + 3H2O 

2KCl + H2SO4 -> K2SO4 + 2HCl

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

2Cl2 + 2Ca(OH)2 -> CaCl2 + Ca(OCl)2 + 2H2O

d, 16HCl + 2KMnO4 -> 5Cl2 + 8H2O + 2KCl + 2MnCl2

3Cl2 + 6KOH (đặc nóng) -> 5KCl + KClO3 + 3H2O

2KCl -> (đpnc) 2K + Cl2

Cl2 + 2NaBr -> 2NaCl + Br2

Br2 + 2NaI -> 2NaBr + I2

Đáp án C

Các phương trình 1,2,4

a. \(2Na+Cl_2\rightarrow2NaCl\)

b. \(C_xH_yO_2+\left(\frac{x+y}{4-1}\right)O_2\rightarrow xCO_2+\frac{y}{2}H_2O\)

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

d. \(NaOH+HCl\rightarrow NaCl+H_2O\)

e. \(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)

f. \(2Fe+2Cl_2\rightarrow2FeCl_3\)

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

h. \(Fe_2O_3+4CO\rightarrow3Fe+4CO_2\)

i. \(Fe_xO_y+2yHCl\rightarrow xFeCl_{\frac{2y}{x}}+yH_2O\)

k. \(Fe_xO_y+yCO\rightarrow xFe+yCO_2\)

l. \(C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+\left(n+1\right)H_2O\)

a. NaCl → Cl₂ → FeCl₃ → NaCl → HCl → CuCl₂ → AgCl.

2NaCl+2H2O-đp\mn->2NaOH+H2+Cl2

3Cl2+2Fe-to>2FeCl3

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

2NaCl+H2SO4-to>Na2SO4+2HCl

2HCl+CuO->CuCl2+H2O

CuCl2+2AgNO3->2AgCl+Cu(NO3)2

b. CaCO₃ → CaCl₂ → NaCl → NaOH → NaClO → NaCl → Cl₂ → FeCl₃ → AgCl

CaCO3+2HCl->CaCl2+H2O+CO2

CaCl2+Na2CO3->2NaCl+CaCO3

2NaCl+2H2o-đp->2NaOH+Cl2+H2

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

2NaClO-to->2NaCl+O2

2NaCl+2H2O-đp\mn->2NaOH+H2+Cl2

3Cl2+2Fe-to>2FeCl3

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

a) 
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

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

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

\(2NaCl+H_2SO_4\underrightarrow{t^o}Na_2SO_4+2HCl\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\downarrow\)

b)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(H_2+Cl_2\underrightarrow{t^o}2HCl\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

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

\(2AgCl\underrightarrow{as}2Ag+Cl_2\)

\(2NaBr+Cl_2\rightarrow2NaCl+Br_2\)

\(2NaI+Br_2\rightarrow2NaBr+I_2\)

\(Zn+I_2\underrightarrow{H_2O}ZnI_2\)

\(ZnI_2+2NaOH\rightarrow2NaI+Zn\left(OH\right)_2\)

c)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

\(3Cl_2+6KOH\underrightarrow{t^o}5KCl+KClO_3+3H_2O\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(2KCl+H_2SO_4\underrightarrow{t^o}K_2SO_4+2HCl\)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(Ca\left(OH\right)_2+Cl_2\rightarrow CaOCl_2+H_2O\)

bài 1

câu a:

Mn02 + 4HCl --> MnCl2 + Cl2 + 2H20

Cl2 + H2 -->t° ánh sáng 2HCl

4HCl + Mn02 --> MnCl2 + Cl2 + 2H20

Cl2 + 2Na -->t° 2NaCl

2NaCl -->điện phân nóng chảy 2Na + Cl2

câu b/

 2KMnO4 + 16HCl (đ) -> 2KCl + 2MnCl2 + 5Cl2 + 8H2O (Đ.C Cl2)
Cl2 + KOH -> KCl + KClO3 + H2O
2KClO3 -> 2KCl + 3O2 (đ/c khí O2 lớp 8)
2KCl -> 2K + Cl2
Cl2 + H2O ->HCl + HClO
2HCl + Fe -> FeCl2 + H2
2FeCl2 + Cl2 -> 2FeCl3
FeCl3 + NaOH -> Fe(OH)3 + NaCl

câu c/

HCl ---> Cl2 ---> FeCl3 ---> NaCl ---> HCl ---> CuCl2 ---> AgCl
2HCl→Cl2+H2
3Cl2+2Fe→2FeCl3
3NaOH+FeCl3→3NaCl+Fe(OH)3
H2SO4+NaCl→HCl+NaHSO4
CuO+2HCl→2H2O+CuCl2
2AgNO3+CuCl2→2AgCl+Cu(NO3)2

Bài 2:

CTHH: MO

\(n_{MO}=\dfrac{15,3}{M_M+16}\left(mol\right)\)

PTHH: MO + 2HCl --> MCl₂ + H₂O

=> \(n_{MCl_2}=\dfrac{15,3}{M_M+16}\left(mol\right)\)

=> \(\dfrac{15,3}{M_M+16}\left(M_M+71\right)=20,8\)

=> M_M = 137 (g/mol)

=> M là Ba (Bari)

\(n_{BaO}=\dfrac{15,3}{153}=0,1\left(mol\right)\)

PTHH: BaO + 2HCl --> BaCl₂ + H₂O

            0,1-->0,2

=> m_HCl = 0,2.36,5 = 7,3 (g)

=> \(m_{ddHCl}=\dfrac{7,3.100}{18,25}=40\left(g\right)\)

Bài 3: 

a) 

MnO₂ + 4HCl --t^o--> MnCl₂ + Cl₂ + 2H₂O

2NaOH + Cl₂ --> NaCl + NaClO + H₂O

b)

n_NaOH = 0,5.4 = 2 (mol)

\(n_{MnO_2}=\dfrac{69,6}{87}=0,8\left(mol\right)\)

PTHH: MnO₂ + 4HCl --t^o--> MnCl₂ + Cl₂ + 2H₂O

             0,8--------------------------->0,8

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

                1,6<--0,8----->0,8----->0,8

=> \(\left\{{}\begin{matrix}n_{NaOH\left(dư\right)}=2-1,6=0,4\left(mol\right)\\n_{NaCl}=0,8\left(mol\right)\\n_{NaClO}=0,8\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{M\left(NaOHdư\right)}=\dfrac{0,4}{0,5}=0,8M\\C_{M\left(NaCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaClO\right)}=\dfrac{0,8}{0,5}=1,6M\end{matrix}\right.\)

1)\(N^{-3}H_3+Cl_2\rightarrow N^0_2+HCl\)

\(\left\{{}\begin{matrix}1\times|2N^{-3}\rightarrow N_2^0+6e\left(oxihóa\right)\\3\times|Cl_2^0+2e\rightarrow2Cl^{-1}\left(khử\right)\end{matrix}\right.\)

\(\Rightarrow2NH_3+3Cl_2\rightarrow N_2+6HCl\). Cl là chất oxi hóa và N là chất khử

2)\(N^{-3}H_3+O_2^0\rightarrow N^{+2}O+H_2O^{-2}\)

\(\left\{{}\begin{matrix}4\times|N^{-3}\rightarrow N^{+2}+5e\left(oxihóa\right)\\5\times|O_2^0+4e\rightarrow2O^{2-}\left(khử\right)\end{matrix}\right.\)

\(\Rightarrow4NH_3+5O_2\rightarrow4NO+6H_2O\). N là chất oxi hóa và O là chất khử

3)\(Al^0+Fe^{+\frac{8}{3}}_3O_4\rightarrow Al^{+3}_2O_3+Fe^0\)

\(\left\{{}\begin{matrix}4\times|2Al^0\rightarrow Al_2^{+3}+6e\left(oxihóa\right)\\3\times|Fe^{+\frac{8}{3}}_3+8e\rightarrow3Fe^0\left(khử\right)\end{matrix}\right.\)

\(\Rightarrow8Al+3Fe_3O_4\rightarrow4Al_2O_3+9Fe\)

Fe là chất khử và Al là chất oxi hóa