解：設(shè)參加反應(yīng)的??$\mathrm{ CaCO_3} $??的質(zhì)量為x，稀鹽酸中溶質(zhì)的質(zhì)量為y，
生成??$\mathrm{CaCl_2 } $??的質(zhì)量為z。
??$\ \text {CaCO}_3+2\ \text {HCl}\xlongequal[\ \ \ \ \ \ ]{\ \ \ \ \ \ }\ \text {CaCl}_2+\ \text {H}_2\ \text {O}+\ \text {CO}_2↑$??
100 73 111 44
x y z 4.4g
??$\mathrm{ \frac {100 }{ x }=\frac {73 }{ y }=\frac { 111 }{ z }=\frac { 44 }{ 4.4g } }$?? 解得 x = 10 g，y = 7.3 g，z = 11.1 g
(1)碳酸鈣的質(zhì)量分?jǐn)?shù)為 ??$\mathrm{\frac{ 10g}{ 12g}\times 100％≈83.3 ％}$??
(2)稀鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)為 ??$\mathrm{\frac{ 7.3g}{ 50g}\times 100％= 14.6％}$??
(3)反應(yīng)后溶液中溶質(zhì)質(zhì)量分?jǐn)?shù)為??$ \mathrm{\frac{ 11.1g}{ 10g＋50g－4.4g}\times 100％≈ 20.0％}$??
