Mn hộ t theo thứ tự STT; CTHH; Phân loại ;Tên gọi; CT axit ứng ; CT bazơ ứng

Đáp án `+` Giải thích các bước giải:

`49-AlCl_3 -` muối trung hòa `- HCl - Al(OH)_3`

`50-FeBr_2-` muối trung hòa `- HBr - Fe(OH)_2`

`51-NaNO_3-` muối trung hòa `- HNO_3 - NaOH`

`52-K_2SO_4 -` muối trung hòa `-H_2SO_4-KOH`

`53-NaH_2PO_4-` muối axit `-H_3PO_4 - NaOH`

`54-BaCO_3 -` muối trung hòa `-H_2CO_3 - Ba(OH)_2`

`55-KHSO_4-` muối axit `H_2SO_4 - KOH`

`56-CuS-` muối trung hòa `-H_2S - Cu(OH)_2`

`57-FeS-` muối trung hòa `-H_2S-Fe(OH)_2`

`58-MgSO_4-` muối trung hòa `-H_2SO_4-Mg(OH)_2`

\begin{array}{|c|c|c|}\hline \text{STT}&\text{CTHH}&\text{Phân loại}&\text{Tên gọi}&\text{CT Acid tương ứng}&\text{CT base tương ứng}\\\hline\text{49}&\text{$AlCl_3$}&\text{Muối trung hòa}&\text{Nhôm Chlorua}&\text{HCl}&\text{$Al(OH)_3$}\\\hline \text{50}&\text{$FeBr_2$}&\text{Muối trung hòa}&\text{Sắt(II) Bromua}&\text{HBr}&\text{$Fe(OH)_2$}\\\hline \text{51}&\text{$NaNO_2$}&\text{Muối trung hòa}&\text{Sodium nitrit}&\text{$HNO_2$}&\text{NaOH}\\\hline \text{52}&\text{$K_2SO_4$}&\text{Muối trung hòa}&\text{Potassium Sulfate}&\text{$H_2SO_4$}&\text{KOH}\\\hline \text{53}&\text{$NaH_2PO_4$}&\text{Muối acid}&\text{Sodium dihyrogen phosphate}&\text{$H_3PO_4$}&\text{NaOH}\\\hline\text{54}&\text{$BaCO_3$}&\text{Muối trung hòa }&\text{Barium carbonate}&\text{$H_2CO_3$}&\text{$Ba(OH)_2$}\\\hline\text{55}&\text{$KHSO_3$}&\text{Muối acid}&\text{Potassium monohydrogen sulfite}&\text{$H_2SO_3$}&\text{KOH}\\\hline\text{56}&\text{CuS}&\text{Muối trung hòa}&\text{Đồng (II) sulfua}&\text{$H_2S$}&\text{$Cu(OH)_2$}\\\hline\text{57}&\text{FeS}&\text{Muối trung hòa}&\text{Sắt (II) sulfua}&\text{$H_2S$}&\text{$Fe(OH)_2$}\\\hline\text{58}&\text{$MgSO_4$}&\text{Muối trung hòa}&\text{magnesium(II) sulfate}&\text{$H_2SO_4$}&\text{$Mg(OH)_2$}\\\hline\end{array}