C(s) + 2H2(g) ⇒ CH4(g)
initial : a moli b moli
reactioneaza x 2x
la echilibru : (a - x) (b - 2x) x
n gaze = (b-x) moli
x·100/(b-x) = 97 100x = 97b - 97x x = 97b/197
daca : V = 100m³ = (b-x)·22,4 b - x = 100/22,4 = 25/5,6 kmoli
⇒ n CH4 = 0,97·25/5,6 = 4,33kmoli = x
nC = nCH4 mC = 4,33·12 = 51,96kg m carbune = 51,96·100/70 = 74,23kg
nH2 = 2x = 8,66kmoli V H2 = 8,66·22,4 = ≈194m³