над нагретой окисью алюминия пропустили 69 г этилового спирта, в результате чего получено 20 л этилена (н.у.). Каков выход этилена в процентах от теоретического?

над нагретой окисью алюминия пропустили 69 г этилового спирта, в результате чего получено 20 л этилена (н.у.). Каков выход этилена в процентах от теоретического?
- Предмет:
  Химия
- Автор:
  marínnpjf
- 5 лет назад

Ответы 1

C2H5OH + Al2O3=C2H4 + H2OРассчитаем кол-во этилового спиртаМ(C2H5OH)=46 г/мольn(C2H5OH)=m/M=1,5 мольТ.к. n(C2H5OH)=n(C2H4),то V(C2H4)=Vm*VV(C2H4)=22,4л/моль*1,5 моль=33,6 л-практический выходВыход продукта реакции рассчиьаем по формуле q=v(практ)/v(теор)*100%q=20л/33,6*100%=60%
- Автор:
  leopoldorcv0
- 5 лет назад
- 0
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над нагретой окисью алюминия пропустили 69 г этилового спирта, в результате чего получено 20 л этилена (н.у.). Каков выход этилена в процентах от теоретического?

Ответы 1

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