Đáp án:

 4/

a/

$MnO_{2} + 4HCl đ \buildrel{{t^o}}\over\longrightarrow MnCl_{2} + Cl_{2} + 2H_{2}O$

$Cl_{2} + H_{2} \buildrel{{as}}\over\longrightarrow 2HCl$

$2HCl \buildrel{{đpdd}}\over\longrightarrow H_{2} + Cl_{2}$

$3Cl_{2} + 2Fe \buildrel{{t^o}}\over\longrightarrow 2FeCl_{3}$

$FeCl_{3} + 3NaOH → Fe(OH)_{3} + 3NaC$l

$2Fe(OH)_{3} \buildrel{{t^o}}\over\longrightarrow Fe_{2}O_{3} + 3H_{2}O$

b/

$2KMnO_{4} + 16HClđ \buildrel{{t^o}}\over\longrightarrow 2KCl + 2MnCl_{2} +5Cl_{2}+8H_{2}O$

$Cl_{2} + 2Na \buildrel{{t^o}}\over\longrightarrow 2NaCl$

$2NaCl \buildrel{{đpnc}}\over\longrightarrow  2Na + Cl_{2}$

$ Cl_{2} + Cu \buildrel{{t^o}}\over\longrightarrow CuCl_{2} $

$ CuCl_{2} + 2AgNO_{3} →2AgCl + Cu(NO_{3})_{2}$

c/

$2NaCl \buildrel{{đpnc}}\over\longrightarrow 2Na + Cl_{2}$

$Cl_{2} + 2Na \buildrel{{đpnc}}\over\longrightarrow  2NaCl$

$ Cl_{2} + 2NaBr → 2NaCl + Br_{2}$

$Br_{2} + 2NaI →2 NaBr + I_{2}$

$ 3I_{2} +2 Al \buildrel{{t^o}}\over\longrightarrow 2AlI_{3}$

d/

$KMnO_{4} + 16HClđ \buildrel{{t^o}}\over\longrightarrow 2KCl+ 2MnCl_{2} + 5Cl_{2}+8H_{2}O$

$Cl_{2} + 2K \buildrel{{t^o}}\over\longrightarrow 2KCl$

$2KCl(tt) + H_{2}SO_{4}đ \buildrel{{t^o}}\over\longrightarrow K_{2}SO_{4} + 2HCl$

$2HCl \buildrel{{đpdd}}\over\longrightarrow H_{2}+ Cl_{2}$

$Cl_{2}+ 2Na \buildrel{{t^o}}\over\longrightarrow 2NaCl $

$2NaCl + 2H_{2}O \buildrel{{đpdd,cmn}}\over\longrightarrow 2NaOH+ Cl_{2} + H_{2}$

$2NaOH + Cl_{2} → NaClO + NaCl + H_{2}O$

Giải thích các bước giải:

 

Đáp án:câu 5

 

Giải thích các bước giải: