d) Considering only the major species and assuming complete combustion, what is

Chemistry
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Consider the combustion of methanol (CH3OH) in air at an equivalence ratio of 0.85. 

d) Considering only the major species and assuming complete combustion, what is the composition (in mole fractions) of each species in the combustion products?

Nov 30th, 2014

Consider the combustion of methanol (CH3OH) in air at an equivalence ratio of 0.85. 

d) Considering only the major species and assuming complete combustion, what is the composition (in mole fractions) of each species in the combustion products?

Answer:

According to the reaction:

CH3OH + 1.5(O2 + 3.76N2) → CO2 + 2H2O + 5.64N2

                                                   1 mol    2 mol     5.64 mol

So the mol total = 5.64+ 1 + 2 = 8.64 mol

Mole fraction of CO2 = 1 mol / 8.64 mol = 0.115

Mole fraction of H2O = 2 mol / 8.64 mol= 0.231

Mole fraction of N2 = 5.64 / 8.64 mol = 0.652


Nov 30th, 2014

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