解：設參加反應的碳酸鈣的質(zhì)量為?$x。$?
?$\mathrm {CaCO}_3\xlongequal[ ]{高溫}\mathrm {CaO}+\mathrm {CO}_2↑$?
100 44
?$x$? 15.4g
?$\frac {100 }{x }=\frac {44 }{15.4\mathrm {g} }，$?解得?$x=35\mathrm {g}$?
則該石灰石樣品中碳酸鈣的質(zhì)量分數(shù)為?$\frac {35 \mathrm {g}}{50 \mathrm {g}}×100％=70％$?
答：該石灰石樣品中碳酸鈣的質(zhì)量分數(shù)為?$70\%$?
