当前位置:首页学历类中考化学->为了节约林木资源,近几年兴起了石头纸。石头纸可用沿海水产养殖

为了节约林木资源,近几年兴起了石头纸。石头纸可用沿海水产养殖中产生的大量废弃贝壳制得。为了测定某种贝壳中碳酸钙的质量分数,取贝壳25 g,加入200 g的稀盐酸,恰好完全反应,称得反应后物质的总质量为216.2 g(假设贝壳中其他物质不与盐酸反应且不溶于水)。请你计算:

  (1)贝壳中碳酸钙的质量分数;

  (2)反应后所得溶液中溶质的质量分数(精确到0.1%)。

答案:

  解:反应生成CO2的质量:25 g+200 g-216.2 g=8.8 g

  设样品中CaCO3的质量为x,反应后生成CaCl2的质量为y,则

  CaCO3+2HCl===CaCl2+H2O+CO2↑

  100       111     44

  x y 8.8 g

  100x=111y=448.8 g

  x=20 g  y=22.2 g

  (1)贝壳中碳酸钙的质量分数:20 g25 g×100%=80%

  (2)反应后溶液的质量为:216.2 g-(25 g-20 g)=211.2 g。反应后所得溶液中溶质的质量分数为: ×100%=10.5%。

本题解析: 暂无解析
更新时间:2022-04-08 17:46
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