Problem 4.95 Combustion of Propane and Butane Mixture A mixture of propane and b

Question

Problem 4.95 Combustion of Propane and Butane Mixture A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.0 mole% H20. After all the water is removed from the products, the residual gas contains 68.6 mole% CO2 and the balance Oz a. What is the mole percent of propane in the fuel? b It now turns out that the uc mixture may contain not only propane and butane but also other hydrocarbons. The c does not contain oxygen. However, the dry combustion gases stil contain 68.6% carb dioxide. We wish to determine the elemental composition (carbon and hydrogen molar percentages) of the fuel feed. (Hint: Calculate the elemental compositions on an oxygen free basis). What is the mole% of carbon in the fuel

Explanation / Answer

Part a

Basis - Wet product gas = 100 mol

Moles of H2O in we product gas = 47% * 100

= 0.47 * 100 = 47 mol

Moles of Dry product gas = 100 - 47 = 53 mol

Mol % of CO2 = moles of CO2 * 100 / moles of dry product gas

68.6/100 = moles of CO2 / 53

Moles of CO2 = 68.6*53/100 = 36.358 mol

Moles of O2 = 53 - 36.958 = 16.642 mol

Atomic O balance

Moles in O2 inlet = moles of O in H2O in wet product + moles of O in CO2 in dry product + moles of O in O2 in dry product

2*moles of O2 = 47 + 2*36.358 + 2*16.642

moles of O2 = 76.5 mol

Let moles of C3H8 = n1

Moles of C4H10 = n2

Carbon balance

3n1 + 4n2 = 36.358........ Eq1

Hydrogen balance

8n1 + 10n2 = 2*47 = 94......... Eq2

Solve Eq1 and eq2 simultaneously

n1 = 6.2105 mol = moles of C3H8

n2 = 4.4316 mol = moles of C4H10

Mol% of C3H8 = n1*100/(n1 + n2)

= 6.2105*100/(6.2105 + 4.4316)

= 58.36%

Part b

Mol of C = (Moles of CO2) * (1mol C / 1 mol CO2)

= 36.358 mol

Mol of H = (Moles of H2O) * (2mol H / 1 mol H2O)

= 47 * 2 = 94 mol

Mol% of C = (36.358*100) / (36.358 + 94)

= 27.89%

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