Maseni udeo

Submitted by bb123su on Thu, 04/10/2014 - 19:08
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Gasna smesa sadrzi 35 masenih procenata vodonika i 65 masenih % azota.Zapreminski udeo vodonika u smesi je(p i t su const) resenje 0,88

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milicaradenkovic - Thu, 04/10/2014 - 19:51

w(H2)=35 % w(N2)=65 %

To znaci da imas 35 g H2 u 100 g rastvora a 65 g N2 u 100 g rastvora

n(H2)=m/Mr=35/2=17.5 mol

V(H2)=n(H2)*Vm=22.4*17.5=392 l

Vn(N2)=m/Mr=65/28=2.32 mol

V(N2)=n(N2)*Vm=2.32*22.4=52 l

Vu=V(H2)+V(N2)=392+52=444 l

w(H2)=V(H2)/Vu=392/444=0.88