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

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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

，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Fe_xO_y，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

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• 2021年江苏无锡中考化学真题

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