Question

某石灰石的樣品32g（其中含不溶于鹽酸的雜質(zhì)）與200g某稀鹽酸恰好完全反應(yīng)，生成13.2gCO₂．求：（1）該稀鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)；（2）該石灰石樣品中CaCO₃的質(zhì)量分?jǐn)?shù)；（3）反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)．

Answer 1

設(shè)石灰石樣品中含CaCO₃的質(zhì)量為x，鹽酸中含HCl的質(zhì)量為y，反應(yīng)后生成CaCl₂的質(zhì)量為z
CaCO₃+2HCl=CaCl₂+H₂O+CO₂↑
100    73   111      44
x       y    z      13.2g

100

x

=

73

y

=

111

z

=

44

13.2g

解之得 x=30g；y=21.9g；z=33.3g
稀鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)=

21.9g

200g

×100%=10.95%
樣品中CaCO₃的質(zhì)量分?jǐn)?shù)=

30g

32g

×100%=93.8%．
反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)=

33.3g

(30+200-13.2)g

×100%=15.4%
答：（1）稀鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)為10.95%；（2）石灰石樣品中CaCO₃的質(zhì)量分?jǐn)?shù)為93.8%；（3）所得CaCl₂溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為15.4%．