ugljovodonici

Vesna's picture

42. Ako je pri sagorevanju etina dobijeno 300ml gasne smese, koliko je za to utroseno ml kiseonika (isti gasni uslovi)?

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milicaradenkovic's picture

2C2H2+5O2---------------->4CO2+2H2O

gasna smesa je CO2 i H2O

n(gasne smese)=4+2=6

6 mol-----------------------300 ml

5 mol---------------------------x ml

x=300*5/6=250 ml kiseonika

Niki1995's picture

2C2H+ 5O2------->4CO2 + 2H2O
V(gasne smese)=300 ml=0.3 l
n=V/Vn=0.3/22.4=0.013392857 mol
0.013392857mol -------- 6 mol (4 CO2 + 2H2O)
x mol -------------------------- 5 mol
x=0.011160714 mol O2 
V(O2)=n*Vn= 0.011160714* 22.4=0.25 l = 250 ml