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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% H2O. After all the water is removed from the products, the residual gas contains 68.6 mole% CO2 an the balance 02 a. What is the mole percent of propane in the fuel? b. It now turns out that the fuel mixture may contain not only propane and butane but also other hydrocarbons. The fuel does not contain oxygen. However, the dry combustion gases still contain 68.6% carbon 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? SHOW HINT LINK TO TEXT

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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