问题 选择题

取相同体积的KI、Na2S、FeBr2溶液,分别通入足量的Cl2,当反应恰好完成时,消耗Cl2的体积相同(同温、同压下),则KI、Na2S、FeBr2溶液的物质的量浓度之比是(     )

A.2:1:3

B.6:3:2

C.1:1:2

D.3:2:1

答案

答案:B

题目分析:

1、Cl2+2KI=2KCl+I2

2、Cl2+Na2S=S+2NaCl

3、2FeBr2+3Cl2==2FeCl3+2Br2

同温同压下体积相同,则物质的量相同假设均消耗掉1mol氯气,则需要碘化钾2mol,硫化钠1mol,溴化亚铁:2/3mol,所以KI:Na2S:FeBr2=2:1:2/3=6:3:2。故选B。

点评:本题考查物质的有关计算,题目难度不大,本题注意把握物质的量与质量、浓度等物理量的计算公式的运用。氧化性Cl2>Br2>Fe3+

完形填空
完形填空。

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