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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)
3Cl2+6KOH→3H2O+5KCl+KClO3
2KClO3→2KCl+3O2
2KCl+H2SO4→K2SO4+2HCl
16HCl+2KMnO4→5Cl2+8H2O+2KCl+2MnCl2
2Ca(OH)2+2Cl2→2H2O+CaCl2+Ca(ClO)2 e)a. 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
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)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(3Cl_2+6KOH\underrightarrow{t^o}5KCl+KClO_3+3H_2O\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(2KCl+H_2SO_4\underrightarrow{t^o}K_2SO_4+2HCl\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(Ca\left(OH\right)_2+Cl_2\rightarrow CaOCl_2+H_2O\)
2) 2KMnO4 + 16HCl = 2KCl + 2MnCl2 + 5Cl2 + 8H2O
Cl2 + H2 = 2HCl ( điều kiện ánh sáng )
2HCl + Fe = FeCl2 + H2
FeCl2 + 2AgNO3 = 2AgCl + Fe(NO3)2
2AgCl = 2Ag + Cl2
a. KClO3 + 6HCl → 3Cl2 + KCl + 3H2O.
2Fe + 3Cl2 \(\underrightarrow{t^o}\) 2FeCl3.
FeCl3 + 3NaOH \(\rightarrow\) Fe(OH)3 + 3NaCl.
2Fe(OH)3 + 3H2SO4 \(\rightarrow\) Fe2(SO4)3 + 6H2O.
b. KClO3 + 6HCl → 3Cl2 + KCl + 3H2O.
2Na + Cl2 \(\underrightarrow{t^o}\) 2NaCl.
2NaCl + H2SO4 (đặc) \(\underrightarrow{250^oC}\) Na2SO4 + 2HCl.
2HCl + CuO \(\rightarrow\) CuCl2 + H2O.
CuCl2 + 2AgNO3 \(\rightarrow\) 2AgCl + Cu(NO3)2.
2AgCl + Cu \(\rightarrow\) CuCl2 + 2Ag.
Mình làm câu 2 nhé:
Cho thử quỳ tím:
- Quỳ tím chuyển đỏ -> H2SO4, HCl (1)
- Quỳ tím chuyển xanh -> NaOH
- Quỳ tím không đổi màu -> NaCl, NaI, NaBr (2)
Cho lần lượt các chất (1) tác dụng với BaCl2:
- Xuất hiện kết tủa trắng -> H2SO4
BaCl2 + H2SO4 -> BaSO4 + 2HCl
- Không hiện tượng -> HCl
Cho lần lượt các chất (2) tác dụng với AgNO3:
- Kết tủa màu trắng -> AgCl
NaCl + AgNO3 -> AgCl + NaNO3
- Kết tủa màu vàng nhạt -> NaBr
NaBr + AgNO3 -> NaNO3 + AgBr
Kết tủa màu vàng đậm -> NaI
NaI + AgNO3 -> AgI + NaNO3
a_ \(2HCl-^{đpdd}\rightarrow Cl_2+H_2\)
\(\dfrac{3}{2}Cl_2+Fe-^{t^o}\rightarrow FeCl_3\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(NaCl_{\left(r,k\right)}+H_2SO_{4\left(đ,n\right)}-^{t^o}\rightarrow NaHSO_4+HCl\)
\(2HCl+CuO\rightarrow CuCl_2+H_2O\)
\(CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\)
a)
\(MnO_2+4HCl\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\)
\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(2NaCl\left(r\right)+H_2SO_4đ\rightarrow Na_2SO_4+2HCl\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(CuCl_2+2AgNO_3\rightarrow2AgCl+Cu\left(NO_3\right)_2\)
b) \(2KMnO_4+16HClđ\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(Cl_2+H_2\underrightarrow{as}2HCl\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(FeCl_3+3AgNO_3\rightarrow3AgCl+Fe\left(NO_3\right)_3\)
\(2AgCl\underrightarrow{as}2Ag+Cl_2\)
\(Cl_2+2NaBr\rightarrow2NaCl+Br_2\)
\(Br_2+2NaI\rightarrow2NaBr+I_2\)
\(Zn+I_2\underrightarrow{t^o}ZnI_2\)
\(ZnI_2+2NaOH\rightarrow2NaI+Zn\left(OH\right)_2\)
c. \(MnO_2+4HCl\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\)
\(3Cl_2+6KOH\rightarrow3H_2O+5HCl+KClO_3\)
\(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\)
\(2KCl\left(r\right)+H_2SO_4đ\underrightarrow{t^o}K_2SO_4+2HCl\)
\(MnO_2+4HCl\underrightarrow{t^o}MnCl_2+Cl_2+H_2O\)
\(Cl_2+Ca\left(OH\right)_2\rightarrow CaOCl_2+H_2O\)
e. \(2KMnO_4+16HClđ\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\)
\(2KCl+2H_2O\underrightarrow{dpdd}2KOH+H_2+Cl_2\)
\(H_2+Cl_2\underrightarrow{t^o}2HCl\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(2FeCl_2+Cl_2\underrightarrow{t^o}2FeCl_3\)
\(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\)
\(H_2 + Cl_2 \xrightarrow{ánh\ sáng} 2HCl\\ Fe + 2HCl \to FeCl_2 + H_2\\ 2FeCl_2 + Cl_2 \to 2FeCl_3\\ FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl\\ Fe(OH)_3 + 3HCl \to FeCl_3 + 3H_2O\\ 4HCl + MnO_2 \to MnCl_2 + Cl_2 + 2H_2O\\ \)
\(2NaOH + Cl_2 \to NaCl + NaClO + H_2O\\ NaCl + H_2SO_4 \xrightarrow{t^o} NaHSO_4 + HCl\\ CuO + 2HCl \to CuCl_2 + H_2O\\ CuCl_2 + 2AgNO_3 \to Cu(NO_3)_2 + 2AgCl\)
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
=> \(n_{MCl_2}=\dfrac{15,3}{M_M+16}\left(mol\right)\)
=> \(\dfrac{15,3}{M_M+16}\left(M_M+71\right)=20,8\)
=> MM = 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
=> mHCl = 0,2.36,5 = 7,3 (g)
=> \(m_{ddHCl}=\dfrac{7,3.100}{18,25}=40\left(g\right)\)
