Question

Butane (C4H10) burns completely with 110% of theoretical air entering at 74°F, 1 atm, 50% relative humidity. The dry air component can be modeled as 21% O2 and 79% N2 on a molar basis. The combustion products leave at 1 atm. For complete combustion of butane(C4H10) with theoretical amount of air, what is the number of moles of oxygen (O2) per mole of fuel?

Answer 1

Answer: 6.5 moles of oxygen per mole of fuel

Step-by-step explanation:

Since the air consists of 21% oxygen, 79% nitrogen on a molar basis,

equation for the reaction is given as

C4H10 + 02--4CO2+ 5H20

Will now be

C4H10 + a(02 +3.76N2)--4CO2+ 5H20 + 3.76aN2

0.79 / 0.21 = 3.76 = N2/O2 mole ratio ie for every mole of oxygen, we have 3.76moles of nitrogen. Note that nitrogen is an inert gas and will not take place in reaction.

Balancing oxygen on both sides,

ax 2= 4x2 + 5x1

2a=8+5

a=13/2= 6.5 moles of oxygen per mole of fuel

For 110% of theoretical air

Mole of 0₂calculated = 6.5

=1.1x6.5=7.15

= C₄H₁₀ + 7.15{02+ 3.76N₂}-> 4CO₂+ 5H₂0 + 3.76X7.15N₂

= C₄H₁₀ + 7.15{02+ 3.76N₂}-> 4CO₂+ 5H₂0 + 26.884N₂