avem a kmoli CH4 si 1,05a kmoli NH3
a = 60/22,4 = 2,678 kmoli CH4
=> 2,813 kmoli NH3 impuri introduse in reactor
2,678 kmoli CH4 impur ............. 100%
n.pur ........................................... 90%
= 2,41 kmoli CH4 puri
fata de CH4 calculam randamentul de transformare
2,41 kmoli b kg m kg
CH4 + NH3 + 3/2O2 --> HCN + 3H2O
1 17 27
=> m = 2,41x27/1 = 65,075 kg s-ar fi obtinut 100%
100% ................ 65,075 kg
n% ..................... 54
= 83% (rotunjit)
=> b = 2,41x17/1 = 40,97 kg NH3 s-ar fi consumat la n= 100%
40,97 kg NH3 ................... 100%
m.practic ............................. 83%
= 34 kg NH3
