Question

O butelie de gaz contine 200 L de amestec propan si butan (c.n.). Cunoscand ca, amestecul de alcani e echimolar, calculati:

a) volumul de oxigen, c.n., necesar arderii amestecului
b)volumul de CO2 rezultat prin arderea amestecului

Answer 1

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

ν1. 2ν1. ν1. 2ν1

C3H8 + 5O2 -----> 3CO2 + 4H2O + Q

ν2. 5ν2. 3ν2. 4ν2

V=200 l (c.n.) amestec

ν1=ν2 (echimolar)

a) V(O2)=?

b) V(CO2)=?

a) V=ν*22,4=> ν=V/22,4=200/22,4=8,93 mol amestec

ν=ν1+ν2

ν1=ν2

=> ν=2ν1

2ν1=8,93 |:2

ν1=4,465 mol=ν2

1 mol CH4.........2 mol O2

4,465 mol CH4........x

x=8,93 mol O2

1 mol C3H8.......5 mol O2

4,465 mol C3H8......y

y=22,325 mol O2

ν(O2)=x+y=8,93+22,325=31,225 mol O2

V(O2)=ν(O2)*22,4=31,225*22,4

V(O2)=700 l

b) 1 mol CH4.........1 mol CO2

4,465 mol CH4......a

a=4,465 mol CO2

1 mol C3H8.......3 mol CO2

4,465 mol C3H8.....b

b=13,395 mol CO2

ν(CO2)=a+b=4,465+13,395=17,86 mol

V(CO2)=ν(CO2)*22,4=17,86*22,4

V(CO2)=400 l