Question

4,6kg amestec echimolecular de metan si etan este supus arderii cu o cantitate stoechiometrica de oxigen. Calculati volumul de aer (in litri) masurat la 25 grade Celsius, necesar arderii amestecului.

Answer 1

x - moli CH4
y - moli C2H6

x = y

16x + 30y = 4600 <=> 46x = 4600 => x = 100 moli CH4
y = 100 moli C2H6

CH4 + 2O2 => CO2 + 2H2O + Q

1 mol CH4............2×22,4L O2
100 moli CH4.......x = 4480 L O2


C2H6 + 7/2O2 => 2CO2 + 3H2O + Q

1 mol C2H6...................3,5×22,4L O2
100 moli C2H6..............x = 7840L O2

VO2 total = 12320L

n O2 total = 12320/22,4 = 550 moli


pV = nRT => V = nRT/p = 550×0,082×298/1 = 13439,8 L O2

Vaer(20%O2) = 5×VO2 = 13439,8×5 = 67,2 L