Предмет: Химия, автор: Хусь

какой объем воздуха необходим для сгорания 28 л водорода?

Ответы

Автор ответа: melissa17

0

2H2+O2=2H2O
n(H2)=28/22.4=1.25 моль
1.25=x
2 1
x=0.625
n(O2)=0.625 моль
V(O2)=0.625*22.4=14 дм3

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