问题 实验题

针对利用双氧水和二氧化锰制取氧气的实验,请结合如图1所示的常用的仪器和装置。回答有关问题:

(1)配制溶液:若配制60g溶质质量分数为5%的双氧水溶液,需要30%的双氧水溶液    g。在量取水的操作中,当水面接近所需刻度时,应改用     (填仪器名称)加水至所需刻度。

(2)贴标签:将配制好的溶液装入试剂瓶中,请在如图2所示的标签内填入试剂瓶所贴标签的内容。

(3)根据反应原理选仪器:用双氧水和二氧化锰制取氧气,该反应的化学方程式是                。其发生装置和收集装置可选择上图中的    (填序号)组合。

(4)实验步骤:

①检查氧气发生装置的        

②装药品:先往容器中装入固体粉末,再装入液体试剂。

③收集气体,并用带火星的木条来验满。验满的方法是利用氧气具有         的性质。

④回收处理二氧化锰:把氧气发生装置容器内的残留物经         (写操作名称)洗涤、干燥回收二氧化锰。

答案

(1)=10g 50ml 胶头滴管

(2)

(3)2H2O22H2O+O2↑    ABCFI或ACEF I

(4)气密性   支持燃烧(或助燃、帮助燃烧等均可) 过滤

题目分析:(1)根据溶质的质量=溶液的质量×溶质的质量分数,加入溶剂的质量=溶液的质量-30%的双氧水溶液的质量,量取水时接近刻度线时要用胶头滴管定容;若配制60g溶质质量分数为5%的双氧水溶液,需要30%的双氧水溶液是=10g,需要水的质量是60g-10g=50g,合50mL,在量取水的操作中,当水面接近所需刻度时,应改用胶头滴管滴加至所需刻度;(2)据配制的溶液填写标签;配制的溶液是5%的双氧水溶液;(3)双氧水在二氧化锰的催化作用下生成水和氧气,方程式是:2H2O22H2O+O2↑;该反应不需加热,属于固液常温型,故选择装药品的试管,固定试管的铁架台,单孔塞或有分液漏斗的双孔塞,取二氧化锰的药匙,氧气的密度比空气大,用向上排空气法收集,故可选ABCFI或ACEFI组合制取氧气;试管可用来作为少量试剂的反应容器;(4)①实验开始要检查装置的气密性,防止漏气,收集不到气体;

②往容器中装入固体粉末,具体的操作方法是用药匙将二氧化锰粉末送入试管底部,然后将药匙用干净的纸擦干净;

③验满的方法是利用氧气具有支持燃烧的性质,将带火星的木条放于集气瓶口,观察是否复燃;

④双氧水分解生成水,二氧化锰不溶于水,故可把氧气发生装置容器内的残留物经过滤、洗涤、干燥回收二氧化锰。

阅读理解
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