Question

Se da reactia : C6H5 + 5.5 O2 = 2CO2 + 4CO + 3H2O

considerand ca amestecul fractiilor molare initiale contin in total 8 moli C6H5 , calculati necesarul stoechiometric de O2 (in moli si Nm3) si necesarul de aer (21% O2 si 79% N2)

Answer 1

1mol.....5,5moli
C6H6 + 5,5O2 -> 2CO2 + 4CO + 3H2O
8moli....x=?
x=8*5,5=44moli O2
n=V/Vm => VO2=n*Vm=44*22,4=985,6l O2=0,9856Nm³(normal m³ e o notatie veche a metrilor cubi)

985,6l O2..............21%
V aer...................100%
Vaer=985,6*100/21=4693,33l aer