18.6g锌和铁的混合物与足量硫酸反应,共生成标准状况下的氢气6.72L,求反应前锌和铁的质量.
人气：305 ℃ 时间：2019-10-19 15:11:02
解答
标准状况下6.72L氢气的物质的量是0.3mol
那么参与反应的硫酸物质的量也是0.3mol...
所以我们可以假设锌和铁的物质的量分别是xmol和ymol
于是就有
65x + 56y = 18.6
x + y = 0.3
解得x=0.2y=0.1
所以锌和铁的质量分别是13g和5.6g
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