$ 解:設(shè)樣品中碳酸鈣的質(zhì)量為m。$
$ \text{CaCO}_3+2\text{HCl}=\text{CaCl}_2+\text{H}_2\text{O}+\text{CO}_2\uparrow$
100 44 m 8.8g
$ \frac{100}{44}=\frac{m}{8.8\,\text{g}}$
$ m=\frac{100\times8.8\,\text{g}}{44}=20\,\text{g}$
$ 樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)為\frac{20\,\text{g}}{25\,\text{g}}\times100\%=80\%$
$ 答:該石灰石樣品含碳酸鈣的質(zhì)量分?jǐn)?shù)為80\%。$