Question

In the following problems answer is given in bracket, I just want the steps how to get them:

(a). Calculate the % of each DRY product when coal is burned stoichiometrically in air. The mass analysis of the coal is: 80% C; 10% H₂; 5% S; and 5% ash.(76.7%N, 22.5%CO₂; 0.8% SO₂)

(b). Find the air fuel ratio for stoichiometric combustion of Butane by volume. Calculate the percentage of carbon dioxide present in the DRY flue gas if 30% excess air is used. (30.95/1; 10.6%)

Make sure you cover both parts as Chegg Question allows at max 4 sub-parts. I want all the steps to get the correct answer.

Answer 1

Basis : 100 gm of coal

It contains 80%C, means 80 gm of C, 10%H2, i.e 10gm of H2, 5%S, 5 gm of Sulfur

Atomic weights : C =12, H=1, S= 32

Moles : C= 80/12=6.67, H2= 10 and S=5/32=0.15625

Combustion reactions are   C+O2----->CO2, moles of O2 to be used = 6.67 moles

H2+0.5O2---->H2O, moles of O2 to be used = 5moles

S+O2---->SO2, moles of SO2 to be used = 0.15625 moles

total moles of oxygen required = 6.67+5+0.15625=11.83

since air contains 20%O2 and 80%N2, moles of air required = 11.83/0.2 =59.15 moles

moles of products on dry basis : N2=59.15*0.8=47.32, SO2=0.15625, CO2= 6.67 moles

Total moles (dry basis ) = 47.32+0.15625+6.67=54.15

Mole percentages : N2= 100*47.32/54.15=87.39, SO2=100*( 0.15625/54.15= 0.288

and CO2= 100*6.67/54.15=12.31%

b) Basis : 1 mole of butane. The combustion reaction is C4H10+6.5O2----> 4CO2+ 5H2O

1 mole of butane requires 6.5 moles of oxygen. So moles of O2= 6.5 moles

since air contains 20%O2 and 80%N2, moles of air =6.5/0.2= 33.5 moles

Stoichiometric air fuel ratio =33.5/1 =33.5

Air used is 30% excess, moles of air used = 33.5*1.3=43.55 moles, moles of N2= 43.55*0.8 =34.84 moles

moles of oxygen supplied = 43.55*0.2= 8.7 .moles of oxygen used = 8.7-6.5 =2.2 moles

CO2 formed = 4 moles ( from the stoichiometry of the reaction ).

The products on dry basis contain : CO2= 4 moles, O2= 2.2 and N2= 34.84 moles

Percentage of CO2 on dry bais = 100*4/(4+2.2+34.84)=9.74%