解：?$(2\mathrm {)}$?樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)?$\frac {20 \mathrm {g}-4 \mathrm {g}}{20 \mathrm {g}}×100％= 80 ％$?
?$(3\mathrm {)}$?由試題分析50 g稀鹽酸恰好能與石灰石中的6.9 g碳酸鈣完全反應(yīng)
設(shè)第一次參與反應(yīng)的?$\mathrm {HCl}$?質(zhì)量為?$x。$?
?$\mathrm {CaCO}_3+2\mathrm {HCl}\xlongequal[ ]{ }\mathrm {CaCl}_2+\mathrm {H}_2\mathrm {O}+\mathrm {CO}_2↑$?
100 73
6.9g ?$x$
?
?$\frac {100}{6.9 \mathrm {g}}=\frac {73}{x}，$?解得：?$x=5.037 \mathrm {g}$
?
鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)為?$\frac {5.037\mathrm {g}}{50 \mathrm {g}}×100\%=10.074\%$
?
答：鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)為?$10.074\%。$?
