設(shè)碳酸鈣的質(zhì)量是x,該稀鹽酸中溶質(zhì)的質(zhì)量是y,生成CaCl2的質(zhì)量是z.
CaCO3+2HCl=CaCl2+H2O+CO2↑
100 7311144
x yz3.52g
100/x=73/y=111/z=44/3.52
x=8g
y=5.84g
z=8.88g
該大理石的純度是8/10*100%=80%
該稀鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)是5.84/100*100%=5.84%
反應(yīng)后所得溶液的質(zhì)量是8+100-3.52=104.48g,溶質(zhì)質(zhì)量分?jǐn)?shù)是8.88/104.48*100%=8.4%
答：（1）該大理石的純度是80%
（2）該稀鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)是5.84%
（3）完全反應(yīng)后所得溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)是8.4%