已知1molCaCO3分解需要吸收178kJ 热量，1mol焦炭完全燃烧放出39_爱下问搜题

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问题计算题

已知1molCaCO₃分解需要吸收178kJ 热量，1mol焦炭完全燃烧放出393.5kJ ，试回答下列问题

（1）写出碳酸钙分解的热化学反应方程式

（2）试计算如果0.5t CaCO₃煅烧成CaO(s)，在理论上要用焦炭多少千克

答案

（5分）（1）CaCO₃(s)=" CaO(s)+" CO₂(g)，△H="+178kJ/mol" （2）27.14kg

题目分析：（1）1molCaCO₃分解需要吸收178kJ 热量，则该反应的热化学方程式是CaCO₃(s)＝ CaO(s)＋CO₂(g)，△H＝＋178kJ/mol。

（2）0.5t CaCO₃的物质的量是500000g÷100g/mol＝5000mol

实验需要吸收的能量是5000mol×178kJ/mol＝890000kJ

因此理论上要用焦炭的物质的量是890000÷393.5＝2261.8mol

质量是2261.8mol×12g/mol＝27141g＝27.14kg

点评：该题是基础性试题的考查，也是高考中的常见题型和考点。试题基础性强，侧重对学生基础性知识的巩固与训练，有利于培养学生的逻辑推理能力，提高学生分析问题、解决问题的能力，也有助于培养学生的规范答题能力。

完形填空

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单项选择题

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