2 câu trả lời
Đáp án:
$Br_{2}$+$H_{2}$ → $2HBr_{}$
$HBr_{}$+ $NaOH_{}$→ $NaBr_{}$+$H_{2}O$
$NaBr_{}$+$HCl_{}$ →$NaCl_{}$+$HBr_{}$
$NaCl_{}$+$H_{2}O$→ $NaOH_{}$ +$H_{2}$+ $Cl_{2}$(điện phân có màng ngăn)
$\text{PTHH:}$
$\text{Br$_{2}$+H$_{2}$$\xrightarrow[]{t^o}$2HBr}$
$\text{NaOH+HBr→NaBr+H$_{2}$O}$
$\text{2NaBr+Cl$_{2}$→2NaCl+Br$_{2}$}$
$\text{2NaCl+H$_{2}$O$\xrightarrow[có màng ngăn]{đpdd}$2NaOH+Cl$_{2}$+H$_{2}$}$
