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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+。

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单项选择题

车轮踏面擦伤，如果长度大于60mm时，则应按不应大于（）的速度运行到前方站停运。

A、160Km/h

B、120Km/h

C、80Km/h

D、60Km/h
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