Question

Se supun reactiei de ardere 200L amestec CH4 si C2H6 in raporte molar 1:2..
a) Calculati vol. de aer necesar arderii
b) Masa de H2O rezultata din reactie

Answer 1

V   CH4 + 2O2 ⇒ CO2 + 2H2O
2V C2H6 + 3,5O2 ⇒2CO2 + 3H2O
3V = 200 l       V = 200/3 l
a) VO2 = 2V  + 7V = 9V = 9·200/3 = 600 l 
Vaer = 5·600 = 3000 l = 3m³
b) n H2O = 2nCH4 + 3nC2H6 = 2·200/(3·22,4)+ 3·2·200/(3·22,4) = =8·200/(3·22,4)moli 
m = 18·8·200/(3·22,4) = 426,57 g 

Answer 2

n.(niu)=V/V,m
n= 200 l/ 22,4l/mol=8,928mol amestec din care 
n,metan= 8,928/3=2,976mol
n,etan= 2x8,928/3= 5,952mol

din ecuatiile reactiilor
CH4+ 2O2----> CO2+2H2O                 ori 2,976
C2H6 +7/2O2---> 2CO2+ 3H2O          ori 5,952

total moli O2: 2x2,976+3,5x5,952=>26,782mol
V,O2=n,molx22,4l/mol====> 26,782molx22,4l/mol=
V,aer= 5xV,O2=...........................calculeaza!!!

total masa apa:
18g/mol(2x2,976+3x5,952)mol=....................calculeaza !!!!