Question

Consider a 100 mol mixture that is 69.0% methane (CH4) and 31.0% ethane (C2H6). To this mixture is added 20.0% excess air. Of the methane present, 90.40% reacts, 92.30% of which forms carbon dioxide (CO2), and the balance forms carbon monoxide (CO). Of the ethane present, 86.0% reacts, 92.30% of which forms carbon dioxide, and the balance forms carbon monoxide.

a) What is the theoretical amount of oxygen required for the fuel mixture?

b) What amount of air is added to the fuel mixture?

c) How many moles of methane are present in the product gas?

d) How many moles of ethane are present in the product gas?

e) How many moles of carbon dioxide are present in the product gas?

f) How many moles of carbon monoxide are present in the product gas?

g) How many moles of water vapor are present in the product gas?

h) How many moles of oxygen are present in the product gas?

i) How many moles of nitrogen are present in the product gas?

Answer 1

Methane combustion reactions

CH4 + 2O2 = CO2 + 2H2O

CH4 + 1.5O2 = CO + 2H2O

ethane combustion reactions

C2H6 + 3.5O2 = 2CO2 + 3H2O

C2H6 + 2.5O2 = 2CO + 3H2O

At inlet

Moles of CH4 = 100 x 0.69 = 69 mol

Moles of C2H6 = 100 x 0.31 = 31 mol

CH4 reacted = 0.904 x 69 = 62.376 mol

CO2 produced from CH4 = 0.923 x 62.376 = 57.573 mol

CO produced from CH4 = (1 - 0.923) x 62.376 = 4.803 mol

C2H6 reacted = 0.86 x 31 = 26.66 mol

CO2 produced from C2H6 = 0.923 x 26.66 x 2 = 49.22 mol

CO produced from C2H6 = (1 - 0.923) x 26.66 x 2 = 4.10 mol

Part a

O2 required for CH4 combustion = 3.5 x 69 = 241.5 mol

O2 required for C2H6 combustion = 6 x 31 = 186 mol

theoretical amount of oxygen required for the fuel mixture

= 241.5 + 186 = 427.5 mol

Part b

Air fed = 1.2 x 427.5 = 513 mol

Air added = 513 - 427.5 = 85.5 mol

Part c

moles of methane are present in the product gas

= moles of CH4 at inlet - moles of CH4 reacted

= 69 - 62.376

= 6.624 mol

Part d

moles of ethane are present in the product gas

= moles of C2H6 at inlet - moles of C2H6 reacted

= 31 - 26.66

= 4.34 mol

Part e

moles of carbon dioxide are present in the product gas

= CO2 produced from CH4 + CO2 produced from C2H6

= 57.573 + 49.22

= 106.793 mol

Part F

moles of carbon dioxide are present in the product gas

= CO produced from CH4 + CO produced from C2H6

= 4.803 + 4.10

= 8.903 mol