Question

varianta 92) 4 si varianta 91)2

Answer 1

V91. 2) Scriem ecuatia pe caz general:
CnH2n + HBr -> CnH2n+1Br
14n+81 g compus...........100%
80g Br............................48,48%
14n+81=8000/48,48=165g
14n=165-81=84 => n=6
alchena: C6H12
V92. 4) amestec echimolecular => nr egali de moli
namestec=Vamestec/Vm=56/22,4=2,5moli
deoarece amestecul e echimolecular => avem 2,5moli propan si 2,5 moli butan
1mol.....................3*44g
C3H8 + 5O2 -> 3CO2 + 4H2O
2,5moli..................330g
1mol................................4*44g
C4H10 + (13/2)O2 -> 4CO2 + 5H2O
2,5moli............................440g
mCO2=770g