问题 推断题

(4分)已知A、B、C、D、E、F六种物质间有如下的转变关系(①②③发生的都是化合反应),其中A、B、C在通常情况下是三种无色无味的气体,且C为化合物,D是一种无色的液体氧化物,F是一种能使紫色石蕊试液变红的物质。

请回答:写出下列变化的化学方程式:

① A + B → D : ______________________________________     

② B + C → E : ______________________________________      

③ D + E → F :                                              

④ F → D + E :                                              

答案

(1)2H2+O22H2O

(2)2CO+O22CO2

(3)H2O+CO2H2CO3

(4)H2CO3H2O+CO2

由图示可以看出,A与B能燃烧,C与B也能燃烧,所以我们可以推断B为氧气.又因D是一种无色的液体氧化物,可推断D为水.所以我们可确定A为氢气.又因D与E生成F是一种能使紫色石蕊试液变红的物质.所以我们可推断E这二氧化碳,且C为化合物,所以C只能是一氧化碳,F为碳酸,碳酸易分解又可生成二氧化碳和水,依据物质及条件可写出相应的方程式.

故答案为:(1)2H2+O22H2O

(2)2CO+O22CO2

(3)H2O+CO2H2CO3

(4)H2CO3H2O+CO2

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