Em tham khảo nha:

\(\begin{array}{l}
a)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
2FeC{l_2} + C{l_2} \to 2FeC{l_3}\\
b)\\
C{l_2} + 2NaOH \to NaCl + NaClO + {H_2}O\\
3C{l_2} + 6KOH \to 5KCl + KCl{O_3} + 3{H_2}O\\
c)\\
C{l_2} + 2NaBr \to 2NaCl + B{r_2}\\
C{l_2} + 2KBr \to 2KCl + B{r_2}\\
d)\\
C{l_2} + 2NaI \to 2NaCl + {I_2}\\
C{l_2} + 2KI \to 2KCl + {I_2}\\
e)\\
B{r_2} + 2NaI \to 2NaBr + {I_2}\\
B{r_2} + 2KI \to 2KBr + {I_2}
\end{array}\)

a,

$H_2+Cl_2\buildrel{{as}}\over\to 2HCl$

$Cu+Cl_2\buildrel{{t^o}}\over\to CuCl_2$

b,

$Cl_2+H_2O\rightleftharpoons HCl+HClO$

$3Cl_2+6KOH\buildrel{{t^o}}\over\to 5KCl+KClO_3+3H_2O$

c,

$Cl_2+2HBr\to 2HCl+Br_2$

$Cl_2+2NaBr\to 2NaCl+Br_2$

d,

$Cl_2+2NaI\to 2NaCl+I_2$

$Cl_2+2KI\to 2KCl+I_2$

e,

$Br_2+2KI\to 2KBr+I_2$

$Br_2+2NaI\to 2NaBr+I_2$