问题 选择题

NaNO2是一种食品添加剂,具有致癌作用.酸性KMnO4溶液与NaNO2的反应为:MnO4-+NO2-+H+→Mn2++NO3-+H2O(未配平).下列叙述正确的是(  )

A.反应中NO2-是氧化剂

B.反应过程中溶液pH减小

C.1molKMnO4可以消耗2.5molNaNO2

D.该反应可说明氧化性NO3->MnO4-

答案

A、反应中氮元素化合价由NO2-中+3价升高为由NO3-中+5价,NO2-发生氧化反应,是还原剂,故A错误;

B、反应中有氢离子参加,反应氢离子浓度减小,反应过程中溶液pH增大,故B错误;

C、反应中氮元素化合价由NO2-中+3价升高为由NO3-中+5价,锰元素化合价由+7价降低为+2价,令1molKMnO4消耗NaNO2的物质的量为x,根据电子转移守恒,则1mol×(7-2)=(5-3)x,解得x=2.5mol,故C正确;

D、反应中MnO4-是氧化剂,NO3-是氧化产物,所以氧化性NO3-<MnO4-,故D错误.

故选C.

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完形填空。
Dear Li Hua,
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选择题