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    【题目】将a g铁和氧化铁的混合物加入足量的盐酸中，充分反应，固体无剩余。测得参加反应的HCl为0.08 mol，放出标准状况下的气体0.224 L。则下列判断中正确的是( )

    A．原混合物中n(Fe)∶n(Fe2O3)＝2∶1

    B．向溶液中滴入KSCN溶液，显红色

    C．无法计算原混合物的质量

    D．溶液中Fe2＋和Fe3＋的物质的量之比为3∶1

    【答案】A

    【解析】因为Fe3＋和H＋的氧化性强弱顺序为Fe3＋>H＋，所以当产生氢气时，Fe3＋已经全部被还原为Fe2＋，故向溶液中滴入KSCN溶液，溶液不变红色；根据题目给出的数据，

    Fe＋2H＋=Fe2＋＋H2

    n2(HCl)=0.02mol n(H2)=0.01mol

    2Fe3＋＋Fe=3Fe2＋，需要的盐酸n1(HCl)=0.06mol，可计算出n(Fe2O3)＝0.01 mol，参加了个反应的n(Fe)＝0.02 mol，答案选A。

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