設硫酸鈉的質(zhì)量為x,生成氯化鈉的質(zhì)量為y
Na2SO4 + BaCl2 ═ 2 NaCl + BaSO4↓
142 117 233
x y 4.66g
142/x=233/4.66g 解得x=2.84g
117/y=233/4.66g 解得y=2.34g
混合物中硫酸鈉的質(zhì)量分數(shù)=2.84g / 4g × 100% = 71%
反應后溶液的質(zhì)量可根據(jù)質(zhì)量守恒定律求得,
反應后所得溶液質(zhì)量=混合物質(zhì)量+所加水質(zhì)量+所加入氯化鋇溶液質(zhì)量-生成硫酸鋇沉淀質(zhì)量.
所得溶液中溶質(zhì)的質(zhì)量分數(shù)=4g-2.84g+2.34g4g+50.66g+50g-4.66g×100%=3.5%
1、混合物中硫酸鈉的質(zhì)量分數(shù)為71%；
2、恰好完全反應時,所得溶液中溶質(zhì)的質(zhì)量分數(shù)為3.5%．額，第二問呢？設BaCl2質(zhì)量為z Na2SO4 + BaCl2 ═ 2 NaCl + BaSO4↓ ·308233 ·z4.66g 308/z=233/4.66g 解得z=6.46 g 質(zhì)量分數(shù) = 6.16 / 50 = 12.32%