现有A、B、C、D、E五种盐溶液,分别由K+、NH4+、Ag+、Ba2+、Al3+、Cl-、Br

现有A、B、C、D、E五种盐溶液,分别由K+、NH4+、Ag+、Ba2+、Al3+

Cl-、Br-、CO32-、SO42-、NO3- 中的阳离子和阴离子各一种组成(五种盐所含阴、阳离子

各不相同)已知:

①A+B→白色沉淀      ②A+D→白色沉淀        ③A+C→白色沉淀

④D+B→白色沉淀+无色气体    ⑤A溶液中c(H+)=c(OH-)

⑥B溶液中c(H+)=1×10-12mol·L-1   ⑦C、D、E三种溶液的pH<7,

根据以上实验事实可以推出(用分子式表示)

A为___________________;B为__________________;C为___________________;D为__________________;E为___________________。

写出以上①②③④各步反应的离子方程式:

①_____________________________ , ②_____________________________,

③_____________________________,④______________________________。

 

答案

A: BaCl2   B:K2CO3    C: AgNO3   D: Al2(SO4)    E:NH4Br

①Ba2++CO32-==BaCO3↓   ② Ba2++SO42-==BaSO4↓  ③ Ag++Cl-==AgCl↓

④2Al3++3CO32-+3H2O===2Al(OH)3↓+3CO2

解析:略

 

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