Question

calculeaza cantitatea de propan in moli care prin ardere elibereaza 10211,7 Kj.
Se cunosc entalpiile de formare standard: Hf C3H8 = 103,66
Hf CO2
Hf H2O

Answer 1

Reactie: C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)
dH=3HfCO2+4HfH2O-HfC3H8
dH=3*(-393,2)+4*(-241,6)+103,66
dH=-1179,6-966,4+103,66=2042,34kJ/mol
Q=dH/n => n=Q/dH=10211,7/2042,34=5moli propan
MC3H8=3*12+8=44g/mol
n=m/M => m=n*M=5*44=220g propan