AlCl3的物质的量是n=cV=0.1*0.2=0.02mol
AlCl3+3AgNO3=3AgCl↓+Al(NO3)3
1 3
0.02mol n(AgCl)
n(AgCl)=0.06mol
m(AgCl)=nM=0.06*143.5=8.61g
生成的沉淀的质量是8.61g.Wow,这么快啊!~没有疑问的话,麻烦采纳,谢谢!= =蛮 现实的吼没有疑问的话,麻烦采纳,谢谢!如果没有疑问的话,请不要继续追问,谢谢!- -
在200ml0.1mol/L的AlCl3溶液中,滴入足量的硝酸银溶液,生成的沉淀的质量是多少?
在200ml0.1mol/L的AlCl3溶液中,滴入足量的硝酸银溶液,生成的沉淀的质量是多少?
化学人气:875 ℃时间:2019-10-23 08:52:29
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