thực hiện chuỗi phản ứng:
1. MnO2–>Cl2–>FeCl3–>NaCl–>HCl–>CuCl2–>Cu(OH)2–>CuO
2. NaCl–>Cl2–>HCl–>AlCl3–>NaCl–>HCl–>FeCl3–>Fe(OH)3–>Fe2O3
3. HCl–>NaCl–>Cl2–>NaClO–>Br2–>I2–>F2–>Cl2–>F2–>NaF–>HF–>SiF4–>F2–>HF
4. MnO2–>Cl2–>HCl–>NaCl–>Cl2–>Javen–>NaHCO3–>Cl2–>clorua vôi–>CaCO3–>Cl2–>kali clorat–>Cl2
Giải thích các bước giải:
1,
\(\begin{array}{l}
\begin{array}{*{20}{l}}
{4HCl}& + &{Mn{O_2}}& \to &{C{l_2}}& + &{2{H_2}O}& + &{MnC{l_2}}
\end{array}\\
3C{l_2} + 2Fe \to 2FeC{l_3}\\
FeC{l_3} + 3NaOH \to 2NaCl + Fe{(OH)_3}\\
\begin{array}{*{20}{l}}
{{H_2}S{O_4}}& + &{2NaCl}& \to &{2HCl}& + &{N{a_2}S{O_4}}
\end{array}\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
Cu{(OH)_2} \to CuO + {H_2}O
\end{array}\)
2,
\(\begin{array}{l}
NaCl \to Na + \frac{1}{2}C{l_2}\\
{H_2} + C{l_2} \to 2HCl\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
AlC{l_3} + 3NaOH \to 3NaCl + Al{(OH)_3}\\
\begin{array}{*{20}{l}}
{{H_2}S{O_4}}& + &{2NaCl}& \to &{2HCl}& + &{N{a_2}S{O_4}}
\end{array}\\
6HCl + F{e_2}{O_3} \to 2FeC{l_3} + 3{H_2}O\\
FeC{l_3} + 3NaOH \to 3NaCl + Fe{(OH)_3}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O
\end{array}\)
3,
\(\begin{array}{l}
HCl + NaOH \to NaCl + {H_2}O\\
NaCl \to Na + \frac{1}{2}C{l_2}\\
2NaOH + C{l_2} \to {H_2}O + NaCl + NaClO\\
NaClO\; + \;2NaBr\; + \;2HCl\; \to \;3NaCl\; + \;B{r_2}\; + \;{H_2}O\\
2NaI + B{r_2} \to 2NaBr + {I_2}\\
{\rm{2NaF + }}{{\rm{I}}_2} \to 2NaI + {F_2}\\
2AgCl + 2{F_2} \to C{l_2} + 2Ag{F_2}\\
2KF + C{l_2} \to 2KCl + {F_2}\\
2NaCl + {F_2} \to 2NaF + C{l_2}\\
NaF + HCl \to NaCl + HF\\
HF + Si{O_2} \to S{\rm{i}}{{\rm{F}}_4} + {H_2}O\\
{F_2} + {H_2} \to 2HF
\end{array}\)
4,
\(\begin{array}{l}
Mn{O_2} + 4HCl \to C{l_2} + 2{H_2}O + MnC{l_2}\\
C{l_2} + {H_2} \to 2HCl\\
NaOH + HCl \to NaCl + {H_2}O\\
C{l_2} + {H_2}O \to HCl + HClO\\
\begin{array}{*{20}{l}}
{{H_2}O}& + &{NaClO}& + &{C{O_2}}& \to &{NaHC{O_3}}& + &{HClO}
\end{array}\\
2C{l_2} + 2Ca{(OH)_2} \to CaC{l_2} + Ca{(ClO)_2} + 2{H_2}O\\
2Ca{(ClO)_2} + C{O_2} + {H_2}O \to CaC{O_3} + CaC{l_2} + 2HClO\\
\dfrac{1}{2}C{l_2} + K \to KCl\\
KCl \to K + \dfrac{1}{2}C{l_2}
\end{array}\)