zapremina

goranspax's picture

Ako se pri sagorevanju amonijaka oslobodi 600 cm3 azota , koliko mililitara kiseonika je utroseno? Resenje 900cm3

milicaradenkovic's picture

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