有一瓶混有氢氧化钡的氢氧化钠固体，从中取出7.71g样品，加水将其完全溶解后得到50.33g溶液。加入20g硫酸铜溶液后，与之恰好完全反应，生成沉淀2.33g。
求：7.71g样品中氢氧化钡的质量；反应后所得溶液中溶质的质量？
ä½ å¥½ 楼主
解ï¼æ ¹æ®ååºæ¹ç¨å¼
Ba(OH)2+Na2SO4=BaSO4+2NaOH
171............233......80
x..............2.33......y
x=1.71g
y=0.80g
m(NaOH) = 7.71-1.71+0.8 = 6.8g
m(溶液) = 50.33+20-2.33 = 68 g
ååºåæå¾æº¶æ¶²ä¸æº¶è´¨çè´¨éåæ°=(6.8/68)*100%=10%
ä¸æç½åé®æå \(^o^)/~
解ï¼æ ¹æ®ååºæ¹ç¨å¼
Ba(OH)2+Na2SO4=BaSO4+2NaOH
171............233......80
x..............2.33......y
x=1.71g
y=0.80g
m(NaOH) = 7.71-1.71+0.8 = 6.8g
m(溶液) = 50.33+20-2.33 = 68 g
ååºåæå¾æº¶æ¶²ä¸æº¶è´¨çè´¨éåæ°=(6.8/68)*100%=10%
ä¸æç½åé®æå \(^o^)/~
温馨提示：答案为网友推荐，仅供参考
第1个回答 2013-08-13
Ba(OH)2+Na2SO4=BaSO4+2NaOH
171......................233......80
x........................2.33......y
x=1.71g
y=0.80g
m(NaOH) = 7.71-1.71+0.8 = 6.8g
m(溶液) = 50.33+20-2.33 = 68 g
反应后所得溶液中溶质的质量分数=(6.8/68)*100%=10%
171......................233......80
x........................2.33......y
x=1.71g
y=0.80g
m(NaOH) = 7.71-1.71+0.8 = 6.8g
m(溶液) = 50.33+20-2.33 = 68 g
反应后所得溶液中溶质的质量分数=(6.8/68)*100%=10%
第2个回答 2013-08-13
难
