题目内容
取0.04mol KMnO4固体加热一段时间后,收集到a mol单质气体,此时KMnO4的分解率为x,在反应后的残留固体中加入过量的浓盐酸,并加热充分反应,又收集到b mol单质气体,设Mn元素全部以Mn2+存在于反应后的溶液中.
(1)a+b=________(用x表示);
(2)当x=________时,(a+b)取最小值,该最小值为________;
(3)当a+b=0.09时,0.04mol KMnO4加热后所得残留固体物质有________,其物质的量分别是________.
答案:0.1-0.02x;1,0.08;KMnO4,K2MnO4,MnO2,0.02mol,0.01mol,0.01mol
解析:
解析:
|
(1)0.1-0.02x (2)1,0.08 (3)KMnO4,K2MnO4,MnO2,0.02mol,0.01mol,0.01mol |
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