某烃含碳、氢两元素的质量比为6:1,该烃0.1摩尔充分燃烧消耗氧气12.44升,求分子式
人气:469 ℃ 时间:2020-01-28 07:55:24
解答
既然是烃,通式为CxHy,碳、氢两元素的质量比为6:1则12x/y=6/1,y=2x,所以通式又可以写成CxH2x,
CxH2x + 3x/2O2 ----xCO2+xH2O
1mol3x/2mol
0.1mol 12.44L/22.4L/mol
得到3x/2=10*12.44/22.4,解得x=3.7,也就近似4了,所以分子式为C4H8.
(看起来题目数据有问题)
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