试题与答案

(7分)今年我国“世界环境日”的主题是“向污染宣战”。某化工厂排放的废水中含有氢

题型:选择题

题目:

(7分)今年我国“世界环境日”的主题是“向污染宣战”。某化工厂排放的废水中含有氢氧化钾和碳酸钾。现取50g废水样品于锥形瓶中,逐渐加入20%的稀硫酸,到恰好完全反应时,共滴加稀硫酸98g,同时收集到2.2g二氧化碳。求废水样品中氢氧化钾的质量。

答案:

题目分析:已知量:废水样品50g,反应硫酸质量为98g×10%=9.8g;生成二氧化碳为2.2g;反应化学方程式为:K2CO3+H2SO4==K2SO4+H2O+CO2↑ ;2KOH +H2SO4= K2SO4+H2O;未知量:废水样品中氢氧化钾的质量;思路分析:由二氧化碳的质量利用化学方程式可以计算出碳酸钾质量及与碳酸钾反应的硫酸的质量;由硫酸的质量及硫酸与氢氧化钾反应的化学方程式可以计算出氢氧化钾的质量。

解:设废水中的碳酸钾质量为x,与碳酸钾反应的硫酸的质量为y,

K2CO3+H2SO4==K2SO4+H2O+CO2

138    98                 44

x     y                  2.2g

    解得:x=6.8g;  y=4.9g

与氢氧化钾反应的硫酸为98g×10%-4.9g=4.9g

设废水中的氢氧化钾的质量为z

2KOH +H2SO4= K2SO4+2H2O

112     98

z       4.9g

  解得 z=5.6g

答:废水样品中氢氧化钾的质量为5.6g

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