20.0克CuO溶于123.5克溫?zé)岬?0.0% H2SO4溶液.冷卻到20℃時(shí)能從溶液中析出多少克CuSO4?5H2O晶體?(20℃時(shí)CuSO4的溶解度為20.7克/100克水)(銅元素相對(duì)原子質(zhì)量按64計(jì)算)
分析:根據(jù)化學(xué)方程式計(jì)算出生成硫酸銅的質(zhì)量,
根據(jù)
×100%,計(jì)算出CuSO
4?5H
2O晶體中硫酸銅的質(zhì)量分?jǐn)?shù),20℃時(shí)CuSO
4飽和溶液中硫酸銅的質(zhì)量分?jǐn)?shù),
最后根據(jù)硫酸銅的總質(zhì)量不變計(jì)算析出CuSO
4?5H
2O晶體的質(zhì)量.
解答:解:
設(shè):20.0gCuO生成硫酸銅的質(zhì)量為x
CuO+H
2SO
4=CuSO
4+H
2O
80 160
20.0g x
= x=40g
溶液質(zhì)量為:20.0g+122.5g=143.5g
未析出晶體前溶液中CuSO
4的質(zhì)量分?jǐn)?shù)為:
×100%=27.9%
20℃飽和液中CuSO
4的質(zhì)量分?jǐn)?shù)為:
×100%=17.1%
CuSO
4?5H
2O晶體中CuSO
4的質(zhì)量分?jǐn)?shù)為:
×100%=64%
設(shè)20℃時(shí)析出mgCuSO
4?5H
2O,其中含CuSO
4為64%×mg;析出晶體后溶液中CuSO
4為17.1%×(143.5g-mg).
因?yàn)镃uSO
4總質(zhì)量不變,則:64%×mg+17.1%×(143.5g-mg)=27.9%×143.5g
m=33
答:20℃時(shí)析出33gCuSO
4?5H
2O晶體.
點(diǎn)評(píng):本題考查利用化學(xué)方程式進(jìn)行計(jì)算的能力,以及溶質(zhì)的質(zhì)量分?jǐn)?shù)的計(jì)算,難度較大;要熟記和理解溶質(zhì)的質(zhì)量分?jǐn)?shù)及其有關(guān)計(jì)算方法,固體溶解度的概念.