вычислите объем кислорода (н.у.),необходимый для сгорания пропана массой 100г.

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С3H8+5O2=t=>3CO2+4H2O

n(C3H8)=100/44=2.27 моль

n(C3H8)=5n(O2)=2.27*5=11.35 моль

V9O2)=11.35*22.4=254.24 литра