Question

將一定量的石灰石放入燒杯中，加入一定量的稀鹽酸，恰好完全反應(yīng)，產(chǎn)生4.4g氣體，反應(yīng)后所得溶液的溶質(zhì)的質(zhì)量分?jǐn)?shù)為20%．（雜質(zhì)不溶于水，也不參與反應(yīng)）計(jì)算：反應(yīng)前稀鹽酸的溶質(zhì)的質(zhì)量分?jǐn)?shù)．（結(jié)果精確到0.1%）

Answer 1

設(shè)生成CaCl₂的質(zhì)量為x，石灰石中CaCO₃的質(zhì)量為y，參與反應(yīng)的HCl的質(zhì)量為z．
CaCO₃+2HCl=CaCl₂+H₂O+CO₂↑
1007311144
yzx4.4g

111

44

=

x

4.4g

x=11.1g

100

44

=

y

4.4g

y=10g

73

44

=

z

4.4g

z=7.3g
所以反應(yīng)后溶液的質(zhì)量=

11.1g

20%

=55.5g
反應(yīng)前稀鹽酸的溶質(zhì)的質(zhì)量分?jǐn)?shù)=

7.3g

55.5g+4.4g-10g

×100%=14.6%
答：反應(yīng)前稀鹽酸的溶質(zhì)的質(zhì)量分?jǐn)?shù)為14.6%．