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Câu 1:

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

    \(\dfrac{1}{2}Cl_2+K\underrightarrow{t^o}KCl\)

     \(2KCl+2H_2O\xrightarrow[cómàngngăn]{đp}2KOH+Cl_2+H_2\)

     \(Cl_2+H_2O⇌HCl+HClO\)

     \(NaOH+HClO\rightarrow NaClO+H_2O\)

     \(NaClO_{\left(rắn\right)}+HCl\rightarrow NaCl+Cl_2+H_2O\)

     \(2NaCl+2H_2O\xrightarrow[cómàngngăn]{đp}2NaOH+Cl_2+H_2O\)

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

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

    \(Cl_2+H_2\underrightarrow{a/s}2HCl\)

    \(2HCl+Fe\rightarrow FeCl_2+H_2\)

    \(2FeCl_2+Cl_2\rightarrow2FeCl_3\)

    \(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_3\downarrow\)

    \(2KCl+2H_2O\xrightarrow[cmn]{đp}2KOH+Cl_2+H_2\)

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

    \(KClO_3+6HCl\rightarrow KCl+3Cl_2+3H_2O\)

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

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

b)

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

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

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

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

3Cl2+2Fe-->2FeCl3

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

2NaCl+H2SO4-->Na2SO4+2HCl

2HCl+Cuo-->CuCl2+H2O

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

3)

câu c

a)

\(16HCl+2KMnO_4\underrightarrow{^{to}}5Cl_2+8H_2O+2MnCl_2\)

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

\(6HCl+KClO_3\rightarrow3Cl_2+3H_2O+KCl\)

\(3Cl_2+2Fe\rightarrow2FeCl_3\)

\(FeCl_3+3KOH\rightarrow Fe\left(OH\right)_3+3KCl\)

\(KCl+NaOH\rightarrow NaCl+KOH\)

2)

\(2H_2O+CaCl_2\underrightarrow{^{đpmn}}Ca\left(OH\right)_2+Cl_2+2H_2\)

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

\(2HCl+Fe\rightarrow FeCl_2+H_2\)

\(2FeCl_2+Cl_2\rightarrow2FeCl_3\)

\(3Ba\left(OH\right)_2+2FeCl_3\rightarrow3BaCl_2+2Fe\left(OH\right)_3\)

3)

\(4HCl+MnO_2\underrightarrow{^{to}}Cl_2+2H_2O+MnCl_2\)

\(Cl_2+2NaOH\rightarrow H_2O+NaCl+NaClO\)

\(H_2O+NaCl+CO_2\rightarrow NaHCO_3+HClO\)

4)

\(KClO_3+3MnO_2+6NaOH\rightarrow3H_2O+KCl+3Na_2MnO_4\)

\(KCl+AgNO_3\rightarrow AgCl+KNO_3\)

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

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

5)

\(CaOCl_2+MnSO_4+2NaOH\rightarrow H_2O+MnO_2+Na_2SO_4+CaCl_2\)

\(2CaOCl_2+H_2O+CO_2\rightarrow CaCO_3+CaCl_2+2HClO\)

\(CaCO_3\underrightarrow{^{to}}CaO+CO_2\)

\(H_2O+NaClO+CO_2\rightarrow NaHCO_3+HClO\)

6)

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

\(Br_2+2HI\rightarrow I_2+2HBr\)

\(2Ag+I_2\rightarrow2AgI\)

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

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

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

Những phản ứng sau là phản ứng oxi hóa – khử:

a)

\(4HCl+MnO_2\rightarrow MnCl_2+Cl_2+2H_2O\\ 3Cl_2+2Fe\xrightarrow[]{t^o}2FeCl_3\\ FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\\ 2NaCl+H_2SO_{4\left(\text{đ}\right)}\xrightarrow[]{t^o}Na_2SO_4+2HCl\)

b)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ Cl_2+H_2\xrightarrow[]{a/s}2HCl\\ 6HCl+Fe_2O_3\rightarrow2FeCl_3+3H_2O\\ FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\)

c)

\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\\ Cl_2+H_2\xrightarrow[]{a/s}2HCl\\ 2HCl+Fe\rightarrow FeCl_2+H_2\\ FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\)

d)

\(3Cl_2+6KOH_{\left(\text{đ}\text{ặ}c\right)}\xrightarrow[]{t^o}5KCl+KClO_3+3H_2O\\ 2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\\ 2KCl\xrightarrow[]{\text{đ}pnc}2K+Cl_2\\ Cl_2+Ca\xrightarrow[]{t^o}CaCl_2\)

e)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ 3Cl_2+6KOH_{\left(\text{đ}\text{ặ}c\right)}\xrightarrow[]{t^o}5KCl+KClO_3+3H_2O\\ 2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\\ 2KCl\xrightarrow[]{\text{đ}pnc}2K+Cl_2\)

g)

\(2KI+O_3+H_2O\rightarrow2KOH+I_2+O_2\\ I_2+H_2\xrightarrow[]{t^o,p,xt}2HI\\ 2HI+Cl_2\rightarrow2HCl+I_2\\ KOH+HCl\rightarrow KCl+H_2O\)