Question

5,6ml etan si propan c.n. dau prin ardere 30,8 mg CO2. Procentul volumetric de etan din amestec este: A.20%,B. 25% C.30% D. 35% E.40%

Answer 1

CH3-CH3 + 7/2O2 --> 2CO2 + 3H2O + Q

CH3-CH2-CH3 + 5O2 ---> 3CO2 + 4H2O + Q

n=V/Vm=5.6/22.4=0.25mmoli amestec
n=m/M=30.8/44=0.7 mmoli amestec CO2

x+y=0.25  |||(-2)==>-2x-2y=-0.5
2x+3y=0.7       ==>2x+3y=0.7
                                 /     y=0.2 mmoli si x+0.2=0.25==>x=0.25-0.2==>x=0.05 mmoli

n=V/Vm==>V=n*Vm=0.05*22.4=1.12 mL etan

%C2H6=VC2H6/Vamestec*100=1.12/5.6*100=20%