某化学实验小组在实验室用下图装置做CaCO3受热分解的实验．加热

问题问答题

某化学实验小组在实验室用下图装置做CaCO₃受热分解的实验．加热一段时间后，同学们对试管内剩余固体的成分进行探究．请你参与探究．

[提出问题]加热CaCO₃后残留的固体成分是什么？

[作出猜想]

猜想1：CaCO₃完全分解，固体成分为 CaO，则反应的化学方程式为______；

猜想2：CaCO₃未完全分解，固体成分为CaO、CaCO₃；

猜想3：CaCO₃未分解，固体成分为CaCO₃．

[实验与事实]

（1）甲同学取少量残留固体于试管中，加适量水振荡后静置，再滴几滴无色酚酞试液，溶液变______色，手摸试管壁感到______．

（2）乙同学取少量残留固体于试管中，先加适量水振荡后静置，再滴几滴无色酚酞试液，溶液变红．接着加过量稀盐酸，有气泡生成，该反应的化学方程式是______．反应过程中固体逐渐消失．若把产生的气体通入澄清的石灰水，将会观察到的现象是______

______．

[结论与交流]

甲同学认为猜想1成立，他的结论是______（填“合理”或“不合理”）的．你认为猜想______成立．

答案

猜想1：根据”CaCO₃完全分解“和结合我们所学的其分解的方程式可以写出其反应式，注意注明条件“高温”反应物是碳酸钙写在等号的左边，生成物是氧化钙和二氧化碳，二氧化碳后面标上上升符号，反应条件是高温，所以方程式是：CaCO₃

高温

CaO+CO₂↑

[实验与事实]（1）CaO+H₂O=Ca（OH）₂，氢氧化钙属于碱，能使酚酞试液变红，滴加几滴无色酚酞试液，溶液变红；氧化钙与水反应放出大量的热，所以手感觉发烫；

（2）根据碳酸钙能与盐酸反应产生气体，反应物是碳酸钙和盐酸，生成物是氯化钙、水、二氧化碳，用观察法配平，二氧化碳后面标上上升符号，所以方程式是：CaCO₃+2HCl=CaCl₂+H₂O+CO₂↑；二氧化碳与氢氧化钙反应生成碳酸钙沉淀和水，所以石灰水变浑浊；甲同学的实验说明含有氧化钙，乙同学的实验说明含有碳酸钙，所以其中含有氧化钙和碳酸钙，所以猜想2成立．

故答案为：CaCO₃

高温

CaO+CO₂↑

（1）红，发烫（或热感）

（2）CaCO₃+2HCl═CaCl₂+H₂O+CO₂↑；石灰水变浑浊．

不合理；你认为猜想2成立．

完形填空。
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