Question

A pure, low molecular weight hydrocarbon gas is burnt in a furnace giving a flue gas containing

10.8% CO 2 , 3.8% O 2 and the remainder nitrogen. Calculate the atomic ratio, H/C, and from this

the formula of the fuel. Note that this analysis is on a dry basis.

2nd part:

Calculate the analysis of the flue gases resulting from the combustion of ethane with 100%

excess air. Give the results on a dry basis.

Answer 1

Basis : 100 moles of flue gas. It contains 10.8 moles CO2, 3.8 moles O2 and rest is =100-(10.8+3.8)= 85.4 moles Nitrogen.

the air is the source for nitrogen, hence moles of air = 85.4/0.79 ( air contains 79%N2 and 21%O2) = 108 moles

moles of O2= 108*0.21= 22.68 moles

from the reaction of C+O2--->CO2, since 1 mole of C produces 1 mole of oxygen

atoms of Carbon = 10.8*6.023*10²³ atoms

oxygen used for prodcing CO2= 10.8 moles

oxygen remaining = 22.68-10.80=11.88 moles

Oxygen present in th flue gas= 3.8 moles. Oxygen used for combustion of H= 11.88-3.80=8.08

the reaction inovling Hydrogen is H2+0.5O2-->H2O, moles of Hydrogen = 8.08*2= 16.16

atoms of H= 2*.16.36*6.023*10²³ atoms ( since H2 means 2 atoms of hydrogen)=16.16*6.023*10²³ atom

C;H ratio = 10.8 : 16.16 = 1:1.5= 2:3

the formula is C2H3

b) Combustion of ethane is C2H6+ 7/2 O2----> 2CO2 + 3H2O

as per the reaction, 1 mole of C2H5 produces 2 moles of CO2 and 3 mole of H2O while consuming 3.5 moles oxygen.

basis : 1 mole of C2H6, moles of oxygen required = 3.5 moles air contains 79% N2 and 21%O2, moles of air = 3.5/0.21=16.67 moles, moles of air actually suppled= 2*16.67 = 33.34 moles

Products ( dry basis ): CO2= 2 moles , N2= 33.34*0.79= 26.34, Oxygen remaining = 33.34*0.21-3.5 =3.5

total moles = 2+26.34+3.5=31.84

composition : 100* moles/ total moles

CO2= 100*2/31.84 = 6.28, N2= 100*26.34/31.84=82.72 and O2= 11.17%