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51问答网 > 将碳酸钠晶体和碳酸氢钠的混合物ng用酒精灯加热至质量不再改变时，得到的固体物质的质量为5.3g，若将ng该

将碳酸钠晶体和碳酸氢钠的混合物ng用酒精灯加热至质量不再改变时，得到的固体物质的质量为5.3g，若将ng该

2025-03-14 14:41:58

推荐回答（1个）

回答1：

加热时，碳酸钠晶体失去结晶水，碳酸氢钠分解：2NaHCO₃

△

Na₂CO₃+CO₂+H₂O，最后剩余固体5.3g为Na₂CO₃，其物质的量=

5.3g

106g/mol

=0.05mol，根据钠离子守恒可知，ng该混合物中钠离子的物质的量为0.05mol×2=0.1mol，溶于水制成200mL溶液，钠离子的物质的量浓度为

0.1mol

0.2L

=0.5mol．L，
故选A．

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