Hexane (C 6 H 14 ) burns with air (21 % O 2 , 79 % N 2 ) in stoichiometric propo
ID: 2258149 • Letter: H
Question
Hexane (C6H14) burns with air (21 % O2, 79 % N2) in stoichiometric proportions. Write the overall chemical reaction for this situation and determine the mole fractions for the C6H14, O2, N2 in the reactant mixture. Also determine the mole fraction of each species in the product mixture.
Hexane (C6H14) burns with air (21 % O2, 79 % N2) in stoichiometric proportions. Write the overall chemical reaction for this situation and determine the mole fractions for the C6H14, O2, N2 in the reactant mixture. Also determine the mole fraction of each species in the product mixture.Explanation / Answer
2C6H14(hexane) + 13O2 (oxygen) ----> 6CO2 + 14H2O
Hexane does n't react with nitogen.
Let's assume number of moles of hexane be 2 mole.
No. of moles of oxygen be 13 and nitrogen will be (13/29)*79= 35.413 mole.
molefraction of hexane = no.of moles of hexane / total number of moles
mole fraction = 2/2+13+35.413
molefraction of hexane = 2/50.413=0.0396
molefraction of o2= 13/50.413 = 0.25786
Molefraction of N2 = 35.413/50.413 = 0.7024
The product contains co2 and water . It also contains unreacted N2....
Molefraction of CO2 = 6/6+14+35.413=6/55.413=0.10827
Molefraction of H2O = 14/55.413=0.25264
Nolefraction of N2= 35.413/55.413=0.6390