M,CH3OH= 32kg/kmol
n = 320kg/32kg=10kmol alcool
din cele doua reactii se deduce
1kmol alcool se obtine din 1 kmol metan----> e nevoie de 10 kmol metan in conditii normale
pV=nRT 5xV= 10X10³x 0,082 x(273+27)------> V=49,2X10³ L= 49,2m³ daca randamentul este 100%
V= 49,2x100/80x100/80= >76,87m³ metan
Q= 0,640kg x 5330kcal/kg= 3411,2kcal; Q=3411,2x4,18kj =14258,8kj
Pc=14258,8kj/0,480kg= 29706 kj/kg etanol VERIFICA !!!!
