Zadatak

4 NH₃ + 3O₂ ---> 2N₂ + 6H₂O

V(O₂)=600 ML

V(N₂)=?

3O₂ --------------- 2N₂

3mola--------------2mola

96g----------------56g

600ml------------V(N₂)

V(N₂)= (600ml*56g)/96g =350 ml

Ali rešenje je 400ml?

Resenje je sledece

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

V(O2)=600 ml

1 mol--------------22.4dm3

x mol--------------0.6dm3

x=0.6*1/22.4=0.0268 mol kiseonika

n(O2):n(N2)=3:2

n(N2)=n(O2)*2/3=0.0268*2/3=0.0178 mol

1 mol--------------22.4dm3

0.0178 mol-----------xdm3

x=0.0178*22.4/1=0.4 dm3 =400 ml azota