Question

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Answer 1

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