QUESTION 330 marks) CH3 CH2 C CH 2-methylpropene is a typical alpha-olefin which

Question

QUESTION 330 marks) CH3 CH2 C CH 2-methylpropene is a typical alpha-olefin which completely oxidizes to carbon dioxide (CO2) and water (H20) in the presence of air. If 0.07 mole of 2-methylpropene (C4Hs) is completely oxidized with 0.9 mol of air in a combustion chamber what would be: (a) the balanced equation for the oxidation reaction? 4 marks 16 marks (c) the final composition of the system assuming that the air comprises 20.95 mole % of oxygen (b) the initial molar composition of all the species present? (02), 78.09 mole % of nitrogen (N2) and 0.96 mole % of argon (Ar)? [10 marks] [Total 100 marks

Explanation / Answer

Part a

Assuming that 1 mol of air consists 79 mol% N2 and 21 mol% O2

Balanced chemical reaction

C4H8 + 6(O2 + 3.76 N2) = 4CO2 + 4H2O + 22.56 N2

Part b

Moles of C4H8 = 0.07

Moles of air = 0.9

Moles of O2 in the air = 0.21*0.9 = 0.189

Moles of N2 in the air = 0.79*0.9 = 0.711

Total Moles of feed = 0.07 + 0.9 = 0.97

Mol% of C4H8 = moles of C4H8 x 100/total moles

= 0.07*100/0.97 = 7.216%

Mol% of O2 = moles of O2 x 100/total moles

= 0.189*100/0.97 = 19.485%

Mol% of N2 = moles of N2 x 100/total moles

= 0.711*100/0.97 = 73.299%

Part C

If air consists 20.95 mol% O2, 78.09 mol% of N2 and 0.96 mol% of Ar.

Moles of air = 0.9

Moles of O2 in the air = 0.2095*0.9 = 0.18855

Moles of N2 in the air = 0.7809*0.9 = 0.70281

Moles of Ar in the air = 0.0096 x 0.9 = 0.00864

N2 and Ar are inert gases, don't participate in the reaction

From the stoichiometry of the reaction

1 mol of C4H8 reacts with = 6 mol of O2

0.07 mol of C4H8 reacts with = 6 x 0.07 = 0.42 mol of O2

But we have only 0.18855 moles of O2 which is less than 0.42.

O2 is limiting reactant

C4H8 is excess reactant

In the final product stream

Moles of O2 = 0

Moles of C4H8 remain = initial - constumed

= 0.07 - (0.18855/6)

= 0.038573

Moles of CO2 formed = 4*(0.18855/6) = 0.125706

Moles of H2O formed = 4*(0.18855/6) = 0.125706

Moles of N2 = 0.70281

Moles of Ar = 0.00864

Total Moles of final stream = 0.038573 + 0.125706 + 0.125706 + 0.70281 + 0.00864

= 1.001435

Mol% of C4H8 = moles of C4H8 x 100/total moles

= 0.038573*100/1.001435

= 3.8517 %

Mol % of CO2 = 0.125706*100/1.001435 = 12.5525%

Mol % of H2O = 0.125706*100/1.001435 = 12.5525%

Mole% of N2 = 0.70281*100/1.001435 = 70.1803%

Mole% of Ar = 0.00864*100/1.001435 = 0.86276%

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