Em tham khảo nha :

\(\begin{array}{l}
a)\\
C{l_2} + 2Na \to 2NaCl\\
2NaCl + {H_2}S{O_4} \to N{a_2}S{O_4} + 2HCl\\
CuO + 2HCl \to CuCl + {H_2}O\\
CuC{l_2} + 2AgN{O_3} \to 2AgCl + Cu{(N{O_3})_2}\\
b)\\
C{l_2} + {H_2} \to 2HCl\\
HCl + AgN{O_3} \to AgCl + HN{O_3}\\
2AgCl \to 2Ag + C{l_2}\\
C{l_2} + 2NaBr \to 2NaCl + B{r_2}\\
B{r_2} + 2NaI \to 2NaBr + {I_2}\\
c)\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
C{l_2} + 2K \to 2KCl\\
2KCl + {H_2}S{O_4} \to {K_2}S{O_4} + 2HCl\\
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
C{l_2} + Ca{(OH)_2} \to CaOC{l_2} + {H_2}O
\end{array}\)

Đáp án+Giải thích các bước giải:

 xin ctlhn để đem về cho nhóm