Question

稱取NaCl和BaCl₂的固體混合物49.3g，加入167g蒸餾水，完全溶解后向該混合溶液中逐滴加入142g的Na₂SO₄溶液，恰好反應(yīng)生成BaSO₄沉淀的質(zhì)量為23.3g．（提示：BaCl₂+Na₂SO₄=BaSO₄↓+2NaCl）．計算恰好完全反應(yīng)時所得溶液中溶質(zhì)的質(zhì)量分數(shù)是多少？

Answer 1

設(shè)恰好反應(yīng)后所得溶液中氯化鈉的質(zhì)量為x，參加反應(yīng)氯化鋇的質(zhì)量為y
BaCl₂+Na₂SO₄═BaSO₄↓+2NaCl
208           233      117
y            23.3g      x   

208

x

=

233

23.3g

=

117

x

x=11.7g    y=20.8g           
反應(yīng)后生成氯化鈉的總質(zhì)量為11.7g+49.3g-20.8g=40.2g
恰好反應(yīng)后所得溶液中氯化鈉的質(zhì)量分數(shù)為

40.2g

49.3g+167g+142g-23.3g

×100%=12%
答：恰好完全反應(yīng)時所得溶液中溶質(zhì)的質(zhì)量分數(shù)是12g．