纳米铁氧化物复合材料制备及其吸附性能研究

发布时间：2018-04-16 07:14

  本文选题：水热法 + 铁氧化物　； 参考：《重庆理工大学》2017年硕士论文

【摘要】：本文采用低成本铁源、碳源和沉淀剂为原料,通过水热法制备得到Fe_3O_4@C、α-Fe_2O_3/γ-AlOOH和Fe_3O_4三种纳米材料。对所得样品进行了表征,探讨原料配比和反应温度对所得产物的重金属离子Cr(VI)和刚果红吸附性能,主要研究内容如下:(1)以氯化铁、环状糊精和尿素为原料进行水热反应,当环状糊精为3mmol,氯化铁与尿素摩尔比为1:0.75-6时,200℃保温12h,得到Fe_3O_4纯相;摩尔比为1:0.75时生成有无定形碳壳包覆的核壳型Fe_3O_4@C复合材料,比表面积为23.07m~2/g,对Cr(VI)和刚果红的吸附质量为10.06mg/g和88.20mg/g。以硝酸铁、环状糊精和尿素为原料进行水热反应,当尿素为30mmol,硝酸铁与环状糊精摩尔比为1:0.15-1.2时,200℃反应12h,得到Fe_3O_4纯相;摩尔比为1:0.3时的产物Fe_3O_4@C为核壳型结构,比表面积为112.91m~2/g。准二级动力学模型与Freundlich模型能较好的描述该样品对刚果红和Cr(VI)的吸附行为,计算得到对Cr(VI)和刚果红的最大吸附量为33.35mg/g和262.72mg/g。(2)以氯化铁、尿素和葡萄糖为原料进行水热反应,当葡萄糖为2mmol,氯化铁、尿素摩尔比为1:0.75-3时,200℃保温12h,得到纯相Fe_3O_4;摩尔比为1:1.5时产物中存在无定形碳结构,比表面积为39.64m~2/g,对Cr(VI)的吸附质量为12.69mg/g。以硝酸铁、尿素和葡萄糖为原料进行水热反应,当葡萄糖为2mmol,硝酸铁、尿素摩尔比为1:0.75-3时,200℃保温12h,得到Fe_3O_4纯相;摩尔比为1:1.5时对应产物的Cr(VI)吸附质量为21.14mg/g,比表面积为85.70m~2/g。Fe3+和部分还原得到的Fe2+在温和沉淀离子OH-的作用下,反应生成Fe_3O_4颗粒,随后碳源碳化反应的碳沉积在粒子的表面,累积到一定程度厚度不再增加,最终形成核壳结构的Fe_3O_4@C。(3)以硝酸铁、硝酸铝和尿素为原料进行水热反应,研究了铁源/铝源摩尔比与尿素含量对产物物相及吸附性能的影响。当2.5mmol尿素,铁源与铝源摩尔比为1:0.33-4时,180℃保温12h,得到α-Fe_2O_3/γ-AlOOH复合材料;铁源/铝源摩尔比为1:4时,所得样品对刚果红的饱和吸附质量为99.66mg/g。(4)以硝酸铁、尿素和CTAB为原料进行水热反应,研究了硝酸铁含量对产物物相及吸附性能的影响。当CTAB为1.1mmol,硝酸铁、尿素摩尔比为1:0.25-1时,200℃保温12h,得到Fe_3O_4纯相;硝酸铁、尿素摩尔比为1:1时,产物Fe_3O_4的刚果红吸附质量分别为99.17mg/g。
[Abstract]:In this paper, using low-cost Tie Yuan, carbon source and precipitator as raw materials, three kinds of nanomaterials, FeSZ _ 3O _ 4C, 伪 -Fe _ 2O _ 3 / 纬 -AlOOH and Fe_3O_4, were prepared by hydrothermal method.The properties of heavy metal ions CrVI and Congo red adsorbed by raw material ratio and reaction temperature were studied. The main contents of the study were as follows: ferric chloride, cyclodextrin and urea were used as raw materials for hydrothermal reaction.When the cyclodextrin is 3 mmol and the molar ratio of ferric chloride to urea is 1: 0.75-6, the pure phase of Fe_3O_4 is obtained at 200 鈩，

本文编号：1757814