1.設(shè)碳酸鈣的質(zhì)量為X,生成氯化鈣的質(zhì)量為Y1.消耗鹽酸的質(zhì)量為M.
CaCO3 + 2HCl ==== CaCl2 + H2O + CO2↑
100………73…………111……………44
X…………M……………Y1……………0.44g
100/X=73/M=111/Y1=44/0.44g
X=1g
Y1=1.11g
M=0.73g
2.設(shè)氧化鈣的質(zhì)量為Z,生成氯化鈣的質(zhì)量為Y2.
CaO + 2HCl ==== CaCl2 + H2O
56………73…………111
Z……(8.03-0.73)…Y2
56/Z=73/7.3=111/Y2
Z=5.6g
Y2=11.1g
反應(yīng)后所得溶液的溶質(zhì)質(zhì)量分數(shù)為:
(1.11g+11.1g)/(1g+5.6g+80.3g-0.44g)100%=14.12%