在一定温度下,向足量的饱和Na2CO3溶液中加入1.06 g无水Na2CO3,搅拌后

在一定温度下,向足量的饱和Na2CO3溶液中加入1.06 g无水Na2CO3,搅拌后静置,最终所得晶体的质量(    )

A.等于1.06 g                       B.大于1.06 g,而小于2.86 g

C.等于2.86 g                       D.大于2.86 g

答案

D


解析:

向饱和Na2CO3溶液中加入1.06 g无水Na2CO3,发生如下反应:

Na2CO3+10H2O====Na2CO3·10H2O

106             286

1.06g          2.86g

由于反应时消耗了原饱和溶液中的溶剂水,导致又析出部分晶体,所以最终析出晶体的质量大于2.86 g。

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