某白色粉末由碳酸钠、硝酸镁、硫酸铜、氯化钾、氯化铵中的一种或几种组成.为了检验它们所含的物质,做了

2025-03-16 02:33:25
推荐回答(1个)
回答1:

根据①溶于水得无色溶液,由于硫酸铜为蓝色溶液,则一定不存在CuSO4
根据②滴加足量稀盐酸,有气泡产生,产生气体只能为二氧化碳,说明一定存在Na2CO3;继续往反应后的溶液中滴加AgNO3溶液有白色沉淀生成,白色沉淀为氯化银,由于加入了稀盐酸,无法判断原溶液是否含有氯离子,即无法确定是否含有KCl;
根据③滴加氢氧化钠溶液并加热,产生使湿润的红色石蕊试纸变蓝的气体,该气体为氨气,证明原溶液中存在NH4Cl,
所以原固体物质中肯定含有的为:Na2CO3、NH4Cl;肯定没有CuSO4;可能含有KCl;
若要检验氯化钾,可以通过焰色反应进行检验,透过蓝色钴玻璃观察火焰颜色呈紫色,证明存在氯化钾,否则一定不存在氯化钾;
反应③为氢氧化钠与氯化铵的反应,在加热条件下,氯化铵与氢氧化钠反应生成氯化钠、水和氨气,反应的化学方程式为:NH4Cl+NaOH
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NaCl+NH3↑+H2O,
故答案为:Na2CO3、NH4Cl;CuSO4;KCl;焰色反应;透过蓝色钴玻璃观察火焰颜色呈紫色;NH4Cl+NaOH
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NaCl+NH3↑+H2O.

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