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[题目]常温下,碳原子数相同的烷烃和烯烃的混合气体,一定体积完全燃烧生成5体积 CO_2,6体积H_2O(均在150℃、1.01×10~5Pa 下测定),试确定该混合气体的可能组成,烷烃和烯烃的体积比及混合气体的体积。解析:据混烃燃烧生成 CO_2与 H_2O 的体积可得混烃中 nc:n_H=5:(6×2)=5:12,由于混烃的体积未知,这样无法判断混烃组成,可进行如下放缩处理:
[Subject] At room temperature, a mixture of alkanes and olefins with the same number of carbon atoms is burned in a certain volume to produce 5 volumes of CO 2 and 6 volumes of H 2 O (measured at 150° C. and 1.01×10 −5 Pa). The possible composition, the volume ratio of alkanes and olefins and the volume of mixed gas. Analysis: According to the volume of CO 2 and H 2 O produced by the combustion of mixed hydrocarbons, nc:n_H=5:(6×2)=5:12 in mixed hydrocarbons can be obtained. Because the volume of mixed hydrocarbons is unknown, the mixed hydrocarbon composition can not be judged, and the following can be performed. Scaling treatment: