a) CH4 + 1/2O2 --> CH3-OH
0.98×200=196m3 CH4 pur
n=V/Vmolar=196/22.4=8.75 kmoli CH4 pur
1kmol.......0.5kmol......1kmol
8.75kmol...x=4.375kmol....y=8.75kmol
n=m/M==>m=n×M=8.75×32=280kg CH3-OH
b) n=mp/mt×100==> mp=n×mt/100=280×80/100=224kg CH3-OH practic
c) Cp=md/ms×100==>ms=md×100/Cp=224×100/20=1120kg soluție
MCH3-OH=32kg/kmol
problema este din cartea de clasa a X ..pag 99..