Question

A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 3 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, and the partial pressure of each gas assuming that the total pressure inside the vessel following the reaction is 5 atm. Assume all gases are ideal

Answer 1

Answer:

Mol fraction CO2 =  0.176

Mol fraction H2O =  0.235

Mol fraction O2 =  0.588

Partial pressure of CO2 = 0.88 atm

Partial pressure of H2O  = 1.175 atm

Partial pressure of O2 = 2.94 atm

Explanation:

Step 1: Data given

amount of O2 = 3 times the amount needed to completely oxidize propane

Step 2: The balanced equation:

C3H8 + 5 O2 → 3 CO2 + 4 H2O

Step 3: Calculate moles at the start

Suppose there is 1.00 mol of C2H8 to start.

Then there is 1.00 * 5 *3.00 = 15.0 moles of O2 to start.

Step 4: Calculate moles at the equilibrium

After the reaction,1.00 mol of C3H8 is completely consumed

There is 3*1.00 = 3.00 moles of CO2 and 4*1.00 = 4.00 moles H2O moles produced

There will be consumed 5*1.00 = 5.00 moles of O2

There will remain 15.00 - 5.00 = 10.00 moles O2

Step 5: Calculate total moles of gases at the end

Total moles of gases at the end: 3 moles CO2  + 4 moles H2O + 10 moles O2 = 17.0 moles  

Step 6: Calculate mol fraction

Mol fraction CO2 = 3.00 moles / 17.00 moles = 0.176

Mol fraction H2O = 4.00 moles / 17.00 moles = 0.235

Mol fraction O2 = 10.00 moles / 17.00 moles = 0.588

Mol fraction CO2 + mol fraction H2O + mol fraction O2 = 0.176 + 0.235 + 0.588 = 1

Step 7: Calculate partial pressure

Partial pressure of CO2 = 0.176 * 5.00 atm = 0.88 atm

Partial pressure of H2O = 0.235 * 5.00 atm = 1.175 atm

Partial pressure of O2 = 0.588 * 5.00 atm = 2.94 atm

Partial pressure of CO2 + partial pressure of H2O + partial pressure of O2 = 0.88 atm + 1.175 atm + 2.94 atm = 5 atm

Answer 2

Answer:

final mole fraction of O₂ = 58.84% , CO₂ = 17.64% , H₂O = 23.52% .

final partial pressure of O₂ = 2.942 atm , CO₂ = 0.882 atm , H₂O = 1.176 atm .

Explanation:

Assuming that propane is present as a gas , and also that the combustion is complete:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

then taking as a reference propane sample= 1 mol , then

initial moles of O₂ = 3* 5 moles = 15 moles

final moles of O₂= 3* 5 moles - 5 moles = 10 moles

final moles of CO₂ = 3 moles

final moles of H₂O = 4 moles

total number of moles = 10 + 3 + 4 = 17 moles

final mole fraction of O₂ = 10/17 = 0.5884 = 58.84%

final mole fraction of  CO₂ = 3/17 = 0.1764 = 17.64%

final mole fraction of  H₂O = 4/17 = 0.2352 = 23.52%

From Dalton's law for ideal gases , the partial pressure p=P*x then

final partial pressure of O₂ = 5 atm * 10/17 = 2.942 atm

final partial pressure of CO₂ = 5 atm * 3/17 = 0.882 atm

final partial pressure of H₂O = 5 atm * 1/17 = 1.176 atm