Предмет: Химия, автор: mashkabortp96rte

При внутримолекулярной дегидратации спирта пропанола – 1 выделяется газ, объёмом 22,4л. Определить массу спирта, вступившего в реакцию.
Заранее спасибо!

Ответы

Автор ответа: Лера5000

1

H₂C₋CH₂-CH₃⇒H₂C=CH-CH₃+H₂O
|
OH
V(H₂C=CH-CH₃)=22.4 л.
n(H₂C=CH-CH₃)=V(H₂C=CH-CH₃)/Vm=22.4л÷22.4л/дм³=1 моль.
n(пропанола-1)=n(H₂C=CH-CH₃)=1 моль.
m(пропанола-1)=n(пропанола-1)*M(пропанола-1)=1 моль*60 г/моль=60 грамм.
Ответ:m(пропанола-1)=60 грамм.

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