(1)設(shè)生成氯化鈣的質(zhì)量為x
CaO+2HCl=CaCl2+H2O
Ca(OH)2+2HCl=CaCl2+2H2O
CaCO3+2HCl=CaCl2+H2O+CO2↑
73 111
292g×10% x
=29.2g
73 |
111 |
29.2g |
x |
x=44.4g
反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)是
44.4g |
316g |
(2)設(shè)參加反應(yīng)的碳酸鈣固體質(zhì)量為y
CaCO3+2HCl=CaCl2+H2O+CO2↑
100 44
y 6.6g
100 |
44 |
y |
6.6g |
y=15g
答案:
(1)此混合溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)14.0%
(2)a3的值15g