化學(xué)興趣小組為測(cè)定某石灰石樣品中碳酸鈣的質(zhì)量分?jǐn)?shù),取26g石灰石樣品放在燒杯中,然后向其中逐漸加入稀鹽酸,使之與樣品充分反應(yīng),恰好完全反應(yīng)(雜質(zhì)不參加反應(yīng))時(shí),加入稀鹽酸的質(zhì)量為90g,反應(yīng)后燒杯中物質(zhì)的總質(zhì)量為105g.請(qǐng)計(jì)算:
(1)反應(yīng)生成二氧化碳的質(zhì)量
(2)樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)(結(jié)果保留到0.1%)
(1)生成二氧化碳的質(zhì)量=26g+90g-105g=11g;
(2)設(shè):樣品中CaCO
3的質(zhì)量為x,
CaCO
3+2HCl═CaCl
2+H
2O+CO
2↑
100 44
x 11g
= 解得:x=25g
樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)是:
×100%=96.2%
答:(1)反應(yīng)生成二氧化碳的質(zhì)量是11g
(2)樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)是96.2%.