问题 实验题

(4分)某化学课堂围绕“酸碱中和反应”,在老师引导下开展探究活动,请你和同学们一起进行实验探究。

【演示实验】将一定量的稀H2SO4加入到盛有NaOH溶液的小烧杯中。

【知识梳理】该反应的化学方程式为(1)             

【提出问题】实验中未观察到明显现象,部分同学产生了疑问:反应后溶液中溶质是什么呢?

【猜想与假设】

甲组:只有Na2SO4    乙组:有Na2SO4和H2SO4 丙组:有Na2SO4和NaOH         

【收集证据】丙组同学取烧杯中的溶液少量于试管中,滴加几滴酚酞试液溶液,无明显变化,溶液中一定没有(2)            

【表达与交流】为了验证其余猜想,甲组同学的实验方案是取少量试液于试管中,滴加氯化钡溶液,观察是否有白色沉淀产生。你认为他们的实验方案(3)        (填“正确”或“不正确”),你的理由是(4)                

乙组同学取少量试液于试管中,加入(5)______             溶液,振荡,观察到(6)___________________________,证明自己的猜想是正确的。

【反思与评价】为回收硫酸钠,应取剩余试液于烧杯中,加入(7)______ 溶液,充分搅拌,蒸发即可得到纯净的硫酸钠。

答案

(1)H2SO4+2NaOH=NaSO4+2H2O    (2)NaOH    (3)不正确

(4)硫酸和氢氧化钠反应生成硫酸钠,不管溶液中是否含硫酸,无色溶液中都会出现白色沉淀

(5)碳酸钠(或碳酸钾或碳酸氢钠、活泼金属、石蕊等)

(6)气泡      (7)适量的碳酸钠(或氢氧化钠)

题目分析:(1) 将稀H2SO4加入到盛有NaOH溶液中, 反应的化学方程式为: H2SO4+2NaOH=NaSO4+2H2O

(2)由于无色酚酞遇碱溶液变红,而取烧杯中的溶液少量于试管中,滴加几滴酚酞试液溶液,无明显变化,溶液中一没有NaOH

(3) 甲组同学的实验方案是取少量试液于试管中,滴加氯化钡溶液,观察是否有白色沉淀产生。你认为他们的实验方案不正确

(4)理由是:硫酸和氢氧化钠反应生成硫酸钠,不管溶液中是否含硫酸,无色溶液中都会出现白色沉淀

(5)抓住甲乙组同学猜想的不同点,是乙组同学中含有硫酸,而甲组同学的猜想中不含硫酸,所以应抓住酸的化学性质,乙组同学取少量试液于试管中,加入碳酸钠(或碳酸钾或碳酸氢钠、活泼金属、石蕊等),若产生气泡等

(6)为回收硫酸钠,就是相当要除去硫酸钠中的硫酸,抓住除杂质的原则,不能引入新的杂质,应取剩余试液于烧杯中,加入适量的碳酸钠(或氢氧化钠)溶液,充分搅拌,蒸发即可得到纯净的硫酸钠

单项选择题
单项选择题

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A.change position or place

B.take action or do something

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D.persuade somebody to change his attitude