摘要:1.Dm=18.4-16.6=1.8 g.恰好完全反应时:NaHCO3+NaOH¾®Na2CO3+H2O. x/124=1.8/18.x=12.4 < 18.4.即NaOH过量.m过量=18.4-12.4=6.0 g, m反应=40×1.8/18=4.0 g,故NaOH%=(4+6)/18.4=54.3%.

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