Zadatak iz oksida

Submitted by bidzip on Fri, 01/19/2018 - 13:10
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Opšta diskusija

Da li moze pomoc oko ovog zadatka.

Koliko se ml azota i kiseonika dobija potpunim razlagajem 300ml azot(i)oksida,

Hvala unapred

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milicaradenkovic - Wed, 01/24/2018 - 14:19

2N2+O2------------->2N2O

n(N2O)=V(N2O)/Vm=0.3/22.4=0.0134 mol

n(N2)=n(N2O)=0.0134 mol

V(N2)=n(N2)*Vm=0.0134*22.4=0.3 l=300 ml

n(O2)=n(N2O)/2=0.0134/2=0.0067 mol

V(O2)=n(O2)*Vm=0.0067*22.4=0.15 l=150 ml