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:
