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氯及其化合物

> 用15.8g高锰酸钾氧化密度为1.19g/cm 3，质量分数为36.5%的盐酸以制取氯气。反应如下： 2KMnO 4+16HCl=2KCl+2MnCl 2+5Cl 2↑+8H 2O 求：（1）生成Cl 2在标况下的体积；（2）被KMnO 4氧化的盐酸的体积；（3）反应中转移的电子总数。

问题：

发布时间：2024-09-11

用15.8g高锰酸钾氧化密度为1.19g/cm ³，质量分数为36.5%的盐酸以制取氯气。反应如下：
2KMnO ₄+16HCl=2KCl+2MnCl ₂+5Cl ₂↑+8H ₂O
求：（1）生成Cl ₂在标况下的体积；
（2）被KMnO ₄氧化的盐酸的体积；
（3）反应中转移的电子总数。

参考答案：

提问用户：Ld***vH4

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用15.8g高锰酸钾氧化密度为1.19g/cm 3，质量分数为36.5%的盐酸以制取氯气。反应如下： 2KMnO ­4+16HCl=2KCl+2MnCl 2+5Cl 2↑+8H 2O 求：（1）生成Cl 2在标况下的体积； （2）被KMnO 4氧化的盐酸的体积； （3）反应中转移的电子总数。

可能感兴趣的题目：

用15.8g高锰酸钾氧化密度为1.19g/cm ³，质量分数为36.5%的盐酸以制取氯气。反应如下：
2KMnO ₄+16HCl=2KCl+2MnCl ₂+5Cl ₂↑+8H ₂O
求：（1）生成Cl ₂在标况下的体积；
（2）被KMnO ₄氧化的盐酸的体积；
（3）反应中转移的电子总数。