Предмет: Химия, автор: minchenkovan

какой объем кислорода необходим для сжигания 48 г пропана?

Ответы

Автор ответа: Анетка125

1

C3H8+5O2=3CO2+4H2O
n(C3H8)=48/44=1.1 моль
n(С3H8) : n(O2)
1 : 5
n(O2)=5.5 моль
V(O2)=22.4*5.5=123.2 л

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