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Submitted by goranspax on Sun, 06/22/2014 - 22:05
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Ako se pri sagorevanju amonijaka oslobodi 600 cm3 azota , koliko mililitara kiseonika je utroseno? Resenje 900cm3

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milicaradenkovic - Mon, 06/23/2014 - 07:11

4NH3+3O2------------------>2N2+6H2O

n(N2)=V(N2)/Vm=0.6/22.4=0.027 mol

n(O2)=3*n(N2)/2=3*0.027/2=0.0405 mol

V(O2=n(O2)*Vm=0.0405*22.4=0.9 l=900 ml