1 mol过氧化钠与2 mol碳酸氢钠固体混合后,在密闭容器中加热充分反

1 mol过氧化钠与2 mol碳酸氢钠固体混合后,在密闭容器中加热充分反应,排出气体物质后冷却,残留的固体物质是(  )

ANa2CO3                             BNa2O2Na2CO3

CNaOHNa2CO3                       DNa2O2NaOHNa2CO3

答案

解析: 可能发生的反应有2NaHCO3Na2CO3CO2H2O2Na2O22CO2=== 2Na2CO3O2,2Na2O22H2O===4NaOHO2;根据计量关系可知,2 mol NaHCO3生成CO2H2O1 molNa2O2只有1 mol,它先与CO2反应,恰好完全反应生成Na2CO3O2,气体排出后,只剩余Na2CO3

答案: A

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