Question

3. The producer gas made from the coke has the following composition :
CO : 28%, CO2 : 3.5%, O2 : 0.5% and N2 : 68%.
The gas burned with such a quantity of air that the oxygen from air is 20% in excess of the net oxygen required for complete combustion. If the combustion is 98% complete, calculate the weight of the gaseous product formed. Given the oxidation reaction as below:
CO + ½ O2  CO2

Answer 1

basis:100 kmol of producer gas:

contains 28kmol of co,3.5 kmol of co₂,0.5kmol of o₂,68 kmol of N₂

from reaction:1 kmol of CO=0.5 Kmol of O₂

THEORITICAL REQUIREMENT OF O₂=0.5*28=14 Kmol

net O₂ demand by difference=14-0.5=13.5 kmol

as air is used 20% in excess,

O₂in supplied air=120*13.5/100=16.2kmol

N₂in supplied air =16.2*79/21=60.94kmol

total moles of N₂ in product gas=kmol of N₂ in air+kmol of N₂ in gas=60.94+68=128.94kmol

total moles O₂=16.2+0.5=16.7 Kmol

conversion of CO to CO₂ is 98%

CO reacted=0.98*28=27.44 kmol

CO unreacted=28-27.44=0.56kmol

from reaction:1kmol of CO=1kmol of CO₂

C0₂in product gas=3.5+27.44=30.94kmol

O₂ reacted=0.5*27.44=13.72kmol

from reaction:1kmol of CO=1kmol of O₂

O₂unreacted=O₂charged-O₂reacted=16.7-13.72=2.98kmol

component	quantitykmol	molecular weight	quantitykg
CO	0.56	28	15.68
CO2	30.94	44	1361.36
O2	2.98	32	95.36
N2	128.94	28	3610.32
TOTAL	163.42		5082.72

5082.72is the weight of product gas obtained based on 100 kmol of producer gas.we have to find the kg of product

M_avg=avg. molecular weight of producer gas

=x_{co.Mco+Xco2.Mco2+Xo2.Mo2+Xn2.Mn2}

_{0.28*28+0.035*44+0.005*32+0.68*28=28.58kg}

_{amount of 100kmol producer gas which is the basis of calculation in weight units}

_{=28.58*100=2858kg}

_{weight of gaseous product formed per kg of producer gas=}

5082.72/2858*100=177.84kg