向0.1mol\L的AL2(SO4)3溶液100mL中加入0.5mol\LNaOH溶液得到1.17沉淀解析:NAOH溶液的体积可能:A.130ML或90ML (1)AI2(SO4)3 0.1mol(2)AI3+ + 3OH- ==AI(OH)3 n(AI3+)==1.17/78 ==0.015mol n(NaOH)==0.015*3 == 0.045mol V(NaOH)==0.045/0.5==0.
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![向0.1mol\L的AL2(SO4)3溶液100mL中加入0.5mol\LNaOH溶液得到1.17沉淀解析:NAOH溶液的体积可能:A.130ML或90ML (1)AI2(SO4)3 0.1mol(2)AI3+ + 3OH- ==AI(OH)3 n(AI3+)==1.17/78 ==0.015mol n(NaOH)==0.015*3 == 0.045mol V(NaOH)==0.045/0.5==0.](/uploads/image/z/773966-38-6.jpg?t=%E5%90%910.1mol%5CL%E7%9A%84AL2%28SO4%293%E6%BA%B6%E6%B6%B2100mL%E4%B8%AD%E5%8A%A0%E5%85%A50.5mol%5CLNaOH%E6%BA%B6%E6%B6%B2%E5%BE%97%E5%88%B01.17%E6%B2%89%E6%B7%80%E8%A7%A3%E6%9E%90%EF%BC%9ANAOH%E6%BA%B6%E6%B6%B2%E7%9A%84%E4%BD%93%E7%A7%AF%E5%8F%AF%E8%83%BD%3AA.130ML%E6%88%9690ML+%281%29AI2%28SO4%293+0.1mol%282%29AI3%2B+%2B+3OH-+%3D%3DAI%28OH%293+n%28AI3%2B%29%3D%3D1.17%2F78+%3D%3D0.015mol+n%28NaOH%29%3D%3D0.015%2A3+%3D%3D+0.045mol+V%28NaOH%29%3D%3D0.045%2F0.5%3D%3D0.)
向0.1mol\L的AL2(SO4)3溶液100mL中加入0.5mol\LNaOH溶液得到1.17沉淀解析:NAOH溶液的体积可能:A.130ML或90ML (1)AI2(SO4)3 0.1mol(2)AI3+ + 3OH- ==AI(OH)3 n(AI3+)==1.17/78 ==0.015mol n(NaOH)==0.015*3 == 0.045mol V(NaOH)==0.045/0.5==0.
向0.1mol\L的AL2(SO4)3溶液100mL中加入0.5mol\LNaOH溶液得到1.17沉淀
解析:
NAOH溶液的体积可能:A.130ML或90ML (1)AI2(SO4)3 0.1mol(2)AI3+ + 3OH- ==AI(OH)3 n(AI3+)==1.17/78 ==0.015mol n(NaOH)==0.015*3 == 0.045mol V(NaOH)==0.045/0.5==0.09L(3)AI3+ + 3OH- ==AI(OH)3 n(NaOH)==1.17*3/78 == 0.045mol AI3+ + 4OH- ==AIO2- +2H2O n(NaOH)=(0.02-0.015)*4==0.02mol V(NaOH)=(0.045+0.02)/0.5 ==0.13L
其中,:n(NaOH)=(0.02-0.015)*4==0.02mol没懂,为什么要(0.02-0.015)*4?
向0.1mol\L的AL2(SO4)3溶液100mL中加入0.5mol\LNaOH溶液得到1.17沉淀解析:NAOH溶液的体积可能:A.130ML或90ML (1)AI2(SO4)3 0.1mol(2)AI3+ + 3OH- ==AI(OH)3 n(AI3+)==1.17/78 ==0.015mol n(NaOH)==0.015*3 == 0.045mol V(NaOH)==0.045/0.5==0.
总共0.02mol铝离子,其中0.015mol的铝离子转化为沉淀,剩下的这0.005mol的铝离子转化成偏铝酸根啊
反应式为AL2(SO4)3+8NaOH=2NaAlO2+3Na2SO4+H2O
离子式为Al3+ + 4OH- =AlO2 - + 2H2O