Question

Butane is burned with air. No carbon monoxide is present in the combustion products. 410 + 13 2 ⁄ 2 → 42 + 52 A 90% conversion of butane is achieved and air is 30% in excess. Using a basis of 100 mol/s of butane: a) Draw and label a flowchart of the process described above. (2) b) Determine the molar flow rate of air fed. (5) c) Calculate the molar flow rate of all the exit gases. (10) d) What could have happened in this reaction if the air supplied was less than the theoretical air

Answer 1

Explanation:

2C

4

H

10

+13O

2

→10H

2

O+8CO

2

Pressure before reaction =P

0

2C

4

H

10

+13O

2

→10H

2

O+8CO

2

At t=0 2 moles 13 moles

At t=t(2−2)(13−13)10moles8moles

= 0 mol = 0 mol of H

2

O(g) of CO

2

At constant volume and temperature :→

PV=nRT

n

P

=

V

RT

constant

n

P

= constant

n

1

P

1

=

n

2

P

2

So, As initial pressure was P

0

P

1

=P

0

initial moles n

1

=15 moles (2+13)

final mole n

2

=18 mole (10+8)

Now,

15

P

0

=

18

P

2

⇒P

2

=

5

6

P

0