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将0.72g草酸亚铁(FeC2O4)放在一个可称量的敞口容器中高温焙烧,500~600℃时,容器中的固体质量保持0.4g不变。所得物质的化学式为(  )


  • A.Fe
  • B.FeO
  • C.Fe2O3
  • D.Fe3O4
答案: C
本题解析:

【分析】根据反应前后铁元素的质量不变,确定所得物质中铁元素和氧元素的质量比,进而确定其化学式。

【解答】解:0.72g草酸亚铁中含铁元素的质量为

image.png

,而反应后固体质量为0.4g,不可能是铁单质,结合选项可知,应为铁的氧化物,铁元素和氧元素的质量比为0.28g:(0.4g﹣0.28g)=7:3,设铁的氧化物的化学式为FexOy,56x:16y=7:3,则x:y=2:3,故所得物质的化学式为Fe2O3


故选:C。

【点评】本题难度不大,明确反应前后铁元素的质量不变、化学式的有关计算是正确解答本题的关键。


更新时间:2022-04-14 14:07
纠错

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