Consider a 100 mol mixture that is 72.0% methane (CH4) and 28.0% ethane (C2H6).

Question

Consider a 100 mol mixture that is 72.0% methane (CH4) and 28.0% ethane (C2H6). To this mixture is added 28.0% excess air. Of the methane present, 94.70% reacts, 87.10% of which forms carbon dioxide (CO2), and the balance forms carbon monoxide (CO). Of the ethane present, 88.7% reacts, 87.10% of which forms carbon dioxide, and the balance forms carbon monoxide. What is the theoretical amount of oxygen required for the fuel mixture? Number 309.76 mol on What amount of air is added to the fuel mixture? Number mol air How many moles of methane are present in the product gas? Number mol CH

Explanation / Answer

Basis:100 moles of fuel mixture containing 72% methane and 28% ethane.

Moles of methane in the fuel= 72 and ethane= 28 moles

The complete combustion reaction of methane and ethane can be represented as

CH4+ 2O2---->CO2+2H2O and C2H6+3.5O2------>2CO2+3H2O

The 1st reaction suggest 1 mole of CH4 requires 2 moles of Oxygen to produce 1 mole of CO2 and 2 moles of H2O.

The second reaction suggests 1 mole of C2H6 required 3.5 moles of O2 to produce 2 moles of CO2 and 3 moles of H2O.

1 mole of CH4 requires 2 moles of oxygen for complete combustion. Hence moles of Oxygen required= 72*2= 144 moles

1mole of ethane requires 3.5 moles of Oxygen for complete combustion. Hence moles of Oxygen required= 28*3.5= 98

Total moles of oxygen required for complete combustion= 144+98=242 moles

Air contains 21% O2 and 79% N2. Moles of air required= 242/0.21=1152 moles

But air is supplied at 28% excess. Moles of air supplied= 1.28*1152=1475 moles

Moles of N2 ( inert component)= 1475*0.79=1165, moles of oxygen supplied= 1475*0.21= 310 moles

Out of 72 moles of methane supplied, only 94.7% got reacted. CH4 reacted= 72*0.947=68 moles

Out of the 68 moles of CH4 reacted, 87.1% reacted to produce CO2. Moles of CH4 reacted to produce CO2= 68*0.871=59 moles. Moles of CH4 conveted to CO= 68-59=9moles

From the reaction-1 moles of Oxygen utilized for combustion of 59 moles of CH4= 2*59=118 moles. Incomplete combustion of CH4 can be represented as CH4+ 1.5O2-->CO+2H2O, moles of Oxygen conumsed for formation of CO=9*1.5=13.5 moles

Out of 28 moles of ethane present, 88.7% reacts. So moles of ethane reated= 28*0.887= 25 moles

Moles of ethane reacted to form CO2= 25*0.871=22moles, moles of ethane reacted to form CO= 25-22= 3 moles

Moles of oxygen requied for complte combustion of C2H6 ( vide reaction-2) = 3.5*22=77 moles

Incomplete combustion of ethane can be represented as C2H6+ 2.5O2--->2CO+3H2O

Moles of oxygen required for incomplete combustion of 3 moles of C2H6= 2.5*3= 7.5 moles

Products contains

CH4 ( unreacted)= 72 ( supplied)- 68= 4 moles

C2H6= 28-25=3 moles

CO2= CO2 produced from complete combustion of CH4+ CO2 produced from compelte combustion of C2H6= 59+2*22 = 103 moles

CO= moles of CO produced from incompelte combustion of CH4 + moles of CO prodcued from incomplete combustion of C2H6= 9+3=12 moles

H2O= H2O prodcued from complete combustion of CH4 + H2O produced from complete combustion of C2H6+ H2O produced from incompelte combustion of CH4+ H2O produced from incompelte combustion of C2H6= 59*2+22*3+9*2+3*3= 211 moles

N2= 1165 moles

O2= O2 supplied- O2 consumed= 310-(118+13.5+77+7.5)=94 moles

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