lzyhl
幼苗
共回答了16个问题采纳率:75% 举报
它与碱石灰生成一种能使湿润的红色石蕊试纸变蓝的气体,说明它含有铵根,与25ml0.1mol/l的硫酸溶液完全反应说明0.271g该物质含有0.005mol的铵根.
它与过量的稀硫酸作用,生成的气体全部通入石灰水中,得0.30g白色沉淀.说明它含有碳酸根或碳酸氢根,若只有碳酸根,0.3g碳酸钙含有0.003mol的碳酸根,0.003mol的碳酸根和0.005mol的铵根共0.27g,也就是说假设不成立,即0.271g该白色固体含有0.005mol的铵根,0.001mol的氢离子,0.003mol的碳酸根(包含碳酸根或碳酸氢根中的).现在可以确定0.271g样品含有碳酸氢胺,碳酸胺,因氢离子只在碳酸氢胺中存在,所以0.271g样品含有0.001mol碳酸氢胺,0.002mol碳酸胺.
0.271g该白色固体含有0.005mol的铵根,0.001mol碳酸氢根,0.002mol碳酸根.
1年前
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