A sample of propane (C3H8) is placed in a closed vessel together with an amount

Question

A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 2.15 times the amount needed to completely oxidize the propane to CO2 and H2O at constant temperature. Calculate the mole fraction of each component in the resulting mixture after oxidation, assuming that the H2O is present as a gas (hint: You need to write the balanced combustion reaction of the reaction between propane and O2 to give CO2(g) and H2O(g), then do your freshman chemistry initial, change, equilibrium chart) (Show all work including units)

Explanation / Answer

first we need write the balenced equation

C3H8 + 5 O2 --------> 3 CO2 + 4 H2O

from the problem it is clear that we need 2.15 times the amount needed for the reaction

5moles x 2.15 = 10.75 moles of O2

from the equation limiting reagent is propane

we can find how much product is produced based on mole ratio

C3H8 : H2O = 1:4 meaning we create 4 moles of H2O

C3H8 : CO2 = 1:3 meaning we create 3 moles of CO2

in the vessel we have 10.75 moles of oxygen but five moles were consumed in the reaction

10.75 - 5 = 5.75 moles of oxygen

total moles = 5.75 + 4 + 3 = 12.75 moles

O2 = 5.75/12.75 = 0.45

H2O = 4/12.75 = 0.313

CO2 = 3/12.75 = 0.235

C3H8 is completely consumed.

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