Question

3．為測(cè)定一瓶敞口放置的濃鹽酸的溶質(zhì)質(zhì)量分?jǐn)?shù)，某同學(xué)量取20ml（密度為1.1g/ml）的該鹽酸與大理石充分反應(yīng)，共用去含碳酸鈣85%的大理石7.1g．計(jì)算該鹽酸的溶質(zhì)質(zhì)量分?jǐn)?shù)．（精確到0.1%）

Answer 1

分析 鹽酸溶質(zhì)的質(zhì)量可根據(jù)碳酸鈣的質(zhì)量算出，鹽酸溶液的質(zhì)量可用鹽酸溶液的體積×鹽酸溶液的密度算出．

解答 解：設(shè)該鹽酸中溶質(zhì)的質(zhì)量為x． 
CaCO₃ +2HCl═CaCl₂+H₂O+CO₂↑
 100     73 
7.1g×85% x 
$\frac{100}{73}=\frac{7.1g×85%}{x}$
x=4.4 g
鹽酸的溶質(zhì)質(zhì)量分?jǐn)?shù)=$\frac{4.4g}{20ml×1.1g/ml}×100%$=20.0%
答：鹽酸的溶質(zhì)質(zhì)量分?jǐn)?shù)是20.0%．

點(diǎn)評(píng) 掌握溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算方法是解題的關(guān)鍵．