Question

有3.6克的碳酸钙高温加热一段时间剩余固体中钙元素57.6%求生成氧化钙的质量

Answer 1

3.6gCaCO3中含Ca的质量是:40/100*3.6=1.44g

剩余固体的质量=1.44/57.6%=2.5g

生成CO2的质量：3.6g-2.5g=1.1g

设mCaO=x

CaCO3=高温=CaO+CO2

````````````````````x``1.1g

````````````````````56``44

x/56=1.1g/44

x=1.4

答：生成氧化钙的质量为1.4g