化学兴趣小组的同学为探究某石灰石的纯度(杂质不与酸反应)向6.0克石灰石样品中逐滴加入稀盐酸至不再产气泡为止,共生成二氧化碳气体1.11L(该条件下二氧化碳的密度为1.977g/L)试计算:
(1)求反应生成二氧化碳的质量(结果精确至0.1)
(2)该石灰石样品中碳酸钙的质量分数是多少?(写出计算过程,结果精确至0.1%)
人气:112 ℃ 时间:2019-11-19 00:17:15
解答
CO2的质量:1.11L*1.977g/L=2.2克CaCO3+2HCl=CaCl2+H2O+CO2↑10044X2.2100:X=44:2.2X=5克碳酸钙的质量分数=5/6=83.3%...
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