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a. MnO2 +4HCl→Cl2+2H2O+MnCl2
2Fe+3Cl2→2FeCl3
FeCl3+3NaOH → Fe(OH)3 +3NaCl
Fe(OH)3+3NaCl→ FeCl3 +3NaOH
FeCl3+3AgNO3→3AgCl+Fe(NO3)3
2AgCl→Cl2+2Ag
b. 2KMnO4 +16HCl→ 5Cl2+8H2O+2KCl+2MnCl2
Cl2+H2→2HCl
CuO+2HCl→CuCl2+H2O
CuCl2+Ba(OH)2→ BaCl2+Cu(OH)2
BaCl2+H2SO4→ BaSO4.+2HCl
c.2 NaCl+H2SO4→ 2HCl +Na2SO4
2HCl→ Cl2 +H2
3Cl2+2Fe→ 2FeCl3
FeCl3+3NaOH →3NaCl +Fe(OH)3
2NaCl+2H2O→2NaOH+Cl2+H2
NaOH+HCl→NaCl+H2O
2NaCl→Cl2+2Na
Cl2+Ca →CaCl2
CaCl2+2AgNO3→2AgCl +Ca(NO3)2
2AgCl→2Ag+Cl2
2Na + Cl2 \(\underrightarrow{t}\) 2NaCl
2NaCl + 2H2O \(\underrightarrow{đpmn}\) 2NaOH + Cl2 + H2
Cl2 + 2NaBr ➝ 2NaCl + Br2
Br2 + 2NaI ➝ 2NaBr + I2
3I2 + 2Al \(\underrightarrow{t}\) 2AlI3
\(Cl_2\) + 2Na => 2NaCl (Điều kiện: nhiệt độ)
2NaCl + \(2H_2O\) => \(Cl_2\) + \(H_2\) + NaOH (Điều kiên: điện phân dung dịch có màng ngăn
\(Cl_2\) + 2NaBr => 2NaCl + \(Br_2\)
\(Br_2\) +2NaI => 2NaBr + \(I_2\)
3\(I_2\) + 2Al => 2\(AlI_3\) (Điều kiện: nhiệt độ, xúc tác: \(H_2O\))
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(2Na+Cl_2\rightarrow2NaCl\)
\(2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgCl_2+2AgNO_3\rightarrow2AgCl+Mg\left(NO_3\right)_2\)
\(2AgCl\rightarrow2Ag+Cl_2\)
a/
\(2HCl\underrightarrow{^{đpdd}}H_2+Cl_2\)
\(3Cl_2+2Fe\underrightarrow{^{to}}2FeCl_3\)
\(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\)
\(2NaCl+H_2SO_4\underrightarrow{^{to}}Na_2SO_4+2HCl\)
\(HCl+CuO\rightarrow CuCl_2+H_2O\)
\(CuCl_2+AgNO_3\rightarrow AgCl+Cu\left(NO_3\right)2\)
b/
\(2HCl\underrightarrow{^{đpdd}}H_2+Cl_2\)
\(Cl_2+2Na\underrightarrow{^{to}}2NaCl\)
\(2NaCl+H_2SO_{4_{dac}}\underrightarrow{^{to}}Na_2SO_4+2HCl\)
\(2HCl+Fe\rightarrow FeCl_2+H_2\)
c/
\(MnO_2+4HCl_đ\underrightarrow{^{to}}MnO_2+Cl_2+2H_2O\)
\(Cl_2+2K\underrightarrow{^{to}}2KCl\)
\(2KCl+H_2SO_4\underrightarrow{^{to}}K_2SO_4+2HCl\)
\(2HCl\underrightarrow{^{đpdd}}H_2+Cl_2\)
\(Cl_2+2NaBr\rightarrow2NaCl+Br_2\)
\(Br_2+2NaI\rightarrow NaBr+I_2\)
d/
\(2KMnO_4+16HCl_đ\underrightarrow{^{to}}2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(Cl_2+H_2\underrightarrow{^{as}}2HCl\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(FeCl_3+3AgNO_3\rightarrow3AgCl+Fe\left(NO_3\right)_3\)
Câu 1:
1.
H2 + Cl2 => 2HCl
H2 + HCl => NaCl + H2
2.
2NaCl => 2Na + Cl2
Cl2 + H2 => 2HCl
Mg + HCl => MgCl2 + H2
3.
2NaCl => 2Na + Cl2
3Cl2 + 2Fe => 2FeCl3
Câu 2:
1.
Ta có:
\(\left\{{}\begin{matrix}n_{SO2}=\frac{12,8}{64}=0,2\left(mol\right)\\n_{NaOH}=\frac{1.250}{1000}=0,25\left(mol\right)\end{matrix}\right.\)
PTHH:
\(SO_2+NaOH\rightarrow NaHSO_4\)
x________x _________x____
\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)
y _______2y________ y ________________
Tỉ lệ : \(1\le\frac{n_{NaOH}}{n_{SO2}}\le2\Leftrightarrow1\le\frac{0,25}{0,2}\le2\)
Giải hệ PT:
\(\left\{{}\begin{matrix}x+y=0,2\\x+2y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,05\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{NaHSO3}=0,15.104=15,6\left(g\right)\\m_{Na2SO3}=0,5.126=6,3\left(g\right)\end{matrix}\right.\)
PTHH: \(Na+O_2\rightarrow2Na_2O\\ Al+HNO_3\rightarrow Al\left(NO_3\right)_3+H_2\\ 2Fe+3Cl_2\rightarrow2FeCl_3\\ 4P+5O_2\rightarrow2P_2O_5\)
A: H2
B: HCl
C: FeCl2
D: FeCl3
E: NaOH
F: Fe(OH)3
G: Fe2O3
H: H2O
PTHH:
(1): Cl2 + H2 =(nhiệt)=> 2HCl
(2): 2HCl + Fe ===> FeCl2 + H2
(3): 2FeCl2 + Cl2 ===> 2FeCl3
(4): FeCl3 + 2NaOH ===> Fe(OH)3\(\downarrow\) + 2NaCl
(5): 2Fe(OH)3 =(nhiệt)=> Fe2O3 + 3H2O
(6): Fe2O3 + 3H2 =(nhiệt)=> 2Fe + 3H2O
\(2NaCl\rightarrow2Na+Cl_2\)
\(Cl_2+H_2\rightarrow2HCl\)
\(6HCl+2Fe\left(OH\right)_3\rightarrow2FeCl_3+3H_2O\)