n,CH4:n,C2H6= 3 : 2
-ecuatiile reactiilor de ardere
CH4+ 2O2---> CO2+ 2H2o ori 3a moli
C2H6+7/2O2---> 2CO2+3H2O ori 2a moli
TOTAL moli CO2= 7a
616g reprezinta masa celor 7a moli-barbotati -> 616= 7a X 44-->a=2mol
molCH4=>6
molC2H6=>4 total 10 mol amestec
a. M,mediu= x1M1+x2M2 x1,x2 sunt fractii molare
M,mediu= 6/10x16 + 4/10x 30= 21,6g/mol
d= M,mediu/M,aer
d= 21,6/29=.................................> calculeaza !!!!
b.m= 6molx16g/mol+ 4molx30g/mol=................................calculeaza
c.V= n xVm
V= 10mol x22,4l.mol=..................................calculeaza !!!