Someone please help. I\'m so lost on this and have a quiz on a problem just like

Question

Someone please help. I'm so lost on this and have a quiz on a problem just like this tomorrow. I've read through my book but I cannot seem to get it right!

100 moles of C4H10 undergoes complete combustion in excess air. First, show the degree of freedom analysis for this process. Then calculate the composition (mole fractions) of the stack gas on a wet basis for the following three cases: (1) theoretical air supplied and 100% conversion of C4H10; (2) 20% excess air and 100% conversion. The energy released in the combustion of C4H10 is 49.2 kJ/g. You require 1000 kJ of heat energy to run a boiler. How much CO2 (in g) will you create by combustion of the required C4H10.

Explanation / Answer

air = 1 mol O2 per 3.76 mol N2

C4H10 + 6.5 O2 ----> 4CO2 + 5H2O , hence inlcluding N2 in eq we get

1) balanced eq is   C4H10 + 6.5 O2 + 24.44 N2 -------> 4CO2 + 5H2O +24.44 N2

theoretical air moles = 24.44 + 6.5 = 30.94 moles for 100 % conversion

stack gas on wet basis means H2O here is also gas( vapor form) included,

total moles = 4+5+24.44 = 33.44

mol fraction of CO2 = ( 4/33.44) = 0.1196 , mol fc of H2O = ( 5/33.44) = 0.1495,

mol fc of N2 = ( 24.44/33.44) = 0.73086,

2) when air = 20 % excess i.e 120 % = 1.2 times air , eq is

C4H10 + 6.5 x 1.2 ( O2 + 3.76N2 ) ---> 4CO2 + 5H2O + 1.3 O2 + 29.328 N2

air supplied = 6.5 x1.2 x 4.76 = 37.128 moles

total moles of stack gas = 4+5+1.3+29.328 = 39.628

mol fc of CO2 = (4/39.628) = 0.1 , mol fc of H2O = 5/39.628 = 0.126, mol fc of O2 = 0.0328

mol fc of N2 = 29.328/39.628 = 0.74

3) moles of C4H10 burnt = 1000 / 49.2 = 20.325

hence moles of CO2 created as per eq = 20.325 x 4 = 81.3

moles of H2O = 5 x20.325 = 101.625 ,

mol of N2 = 20.325 x 6.5 x 3.76 = 496.743

total moles = 81.3 + 101.625 + 496.743 = 679.668

mol fc of CO2 = ( 81.3/679.668) = 0.1196, mol fc of H2O = ( 101.625/679.668) = 0.1495

mol fc of N2 = ( 496.743/679.668) = 0.73086

CO2 in gm = moles x mol wt = 81.3 x44 = 3577.235 gm

degree of freedom analysis

number of equations = 1 , number of independant things = 2 ( 2 reactants)

number of unknowns = 2 ( 1 is amount of air initially , 2nd is excess air left )

Number of degrees = N(unknowns) + N(eq) - N(indipendant things)

                               = 2 + 1 -2 = 1

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