Question

石灰厂为了测定一批石灰石样品中碳酸钙的质量分数,取用4g石灰石样品,把20g稀盐酸分4次加入样品中（样品中除碳酸钙外,其

Answer 1

1）该石灰石样品中碳酸钙的质量分数是

(4g-1g)/4g*100%=75%

CaCO3=高温= CaO+CO2↑

100 56

100t*75% x*92%

100/56=100t*75%/(x*92%)

x=45.65t