[题目]设NA为阿伏加德罗常数的值.下列说法错误的是A．18 g H2O中含有的氢原子数为2NAB．钠在空气中燃烧可生成多种氧化物.23 g钠充分燃烧时转移电子数为NAC．17 g氨气中含有的电子数为10NAD．50 mL 12 mol·L1盐酸与足量MnO2共热.转移的电子数为0.3NA 题目和参考答案——青夏教育精英家教网—

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【题目】设NA为阿伏加德罗常数的值。下列说法错误的是

A．18 g H2O中含有的氢原子数为2NA

B．钠在空气中燃烧可生成多种氧化物，23 g钠充分燃烧时转移电子数为NA

C．17 g氨气中含有的电子数为10NA

D．50 mL 12 mol·L1盐酸与足量MnO2共热，转移的电子数为0.3NA

【答案】D

【解析】18 g H2O的物质的量为1 mol，1个水分子中含有2个氢原子，所以18 g H2O中含有的氢原子数为2NA，A正确；钠在反应中只失去一个电子，23 g钠的物质的量为1 mol，即失去NA个电子，B正确；17 g氨气的物质的量是1 mol，1个氨气含有10个电子，1 mol氨气含有的电子数为10NA，C正确；随着反应进行，盐酸浓度变小，稀盐酸与MnO2不反应，故转移电子数小于0.3NA，D错误。

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