Question

Am si eu o problema ce mi-a dat de gandit, dar nu o stiu face si suna cam asa: Un amestec de 61g matan si butan in raport molar 4:1 e supus arderii. Care este volumul de aer necesar procesului de ardere(20%O2 in aer)?

Answer 1

CH4 + 2 O2 = CO2 + 2 H2O
C4H10 + 13/2 O2 = 4 CO2 + 5 H2O
4*16x+58*y=61
x/y=4/1=>x=4y
4*16*4y+58y=61=>314y=61=>y=0.1942 moli
x = 0.777
1 mol CH4..........2 moli O2
0,777 moli CH4........a moli O2
a = 1,554 moli O2
1 mol C4H10.......13/2 moli O2
0.1942 moli C4H10............b moli
b = 1.2623 moli
V O2 = n O2 * 22.4 = 2.8163*22.4 = 63.08 l
V aer = 100/2 * 63.08 = 315.4 l

Answer 2

MCH4=16g/mol
MC4H10=58g/mol

16x+58y=61⇒16*4y+58y=61⇒122y=61⇒y=0.5 moli C4H10
x/y=4/1⇒x=4y=>x=4*0.5=>x=2 moli CH4

1mol...2mol
CH4 + 2O2 ⇒ CO2 + 2H2O + Q
2mol....x=4mol

1mol..........6.5mol
C4H10 + 13/2O2 ---> 4CO2 + 5H2O + Q
0.5mol.........x=3.25mol

nr total moli O2 amestec = 4+3.25 = 7.25 moli O2

n=V/Vm==>V=n*Vm=7.25*22.4=162.4 L O2

Vaer(20%O2)=5*VO2=5*162.4=812 L O2 amestec