A mixture of ethane and ethene occupied 35.5 L at 1.000 bar and 405 K. This
mixture reacted completely with 110.3 g of O2 to produce CO2 and H2O.
What was the composition of the original mixture? Assume ideal gas
behavior
To find the composition of the original mixture, we first need to determine the moles of ethane (C2H6) and ethene (C2H4) in the reactant mixture.
Step 1: Convert temperature from Celsius to Kelvin
The given temperature is 405 K, which is already in Kelvin.
Step 2: Convert pressure from bar to atm
The given pressure is 1.000 bar. Since 1 bar is equal to 0.987 atm, we can convert the pressure by multiplying 1.000 bar by 0.987:
1.000 bar * 0.987 atm/bar = 0.987 atm
Step 3: Calculate the number of moles of oxygen gas (O2)
To find the number of moles of oxygen gas, we can use the ideal gas equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 atm L/mol K)
T = temperature (in Kelvin)
Rearranging the equation to solve for moles (n), we get:
n = PV / RT
Given:
P = 0.987 atm
V = 35.5 L
R = 0.0821 atm L/mol K
T = 405 K
Substituting the values into the equation:
n = (0.987 atm * 35.5 L) / (0.0821 atm L/mol K * 405 K)
n ≈ 1.63 moles of O2
Step 4: Write and balance the chemical equation
The balanced equation for the reaction between O2, ethane (C2H6), and ethene (C2H4) is:
2C2H6 + 7O2 → 4CO2 + 6H2O
or
C2H6 + 3O2 → 2CO2 + 3H2O
Step 5: Calculate the moles of CO2 and H2O produced
From the balanced equation, we can see that for every 2 moles of ethane (C2H6) reacted, 4 moles of CO2 and 6 moles of H2O are produced.
For 1.63 moles of O2, we can use the ratio from the balanced equation to calculate the moles of CO2 and H2O produced:
Moles of CO2 = (1.63 moles O2 * 4 moles CO2) / 7 moles O2
Moles of CO2 ≈ 0.93 moles
Moles of H2O = (1.63 moles O2 * 6 moles H2O) / 7 moles O2
Moles of H2O ≈ 1.39 moles
Step 6: Calculate the moles of ethane (C2H6) and ethene (C2H4) in the original mixture
From the balanced equation, we can conclude that for every 2 moles of ethane (C2H6) reacted, 2 moles of CO2 are produced. Therefore, the number of moles of ethane is the same as the number of moles of CO2 produced.
Moles of ethane (C2H6) ≈ 0.93 moles
Moles of ethene (C2H4) = Moles of O2 - Moles of ethane
Moles of ethene (C2H4) ≈ 1.63 moles - 0.93 moles
Moles of ethene (C2H4) ≈ 0.70 moles
Step 7: Calculate the composition of the original mixture
To find the composition of the original mixture, we need to find the mole fractions of ethane and ethene.
Mole fraction of ethane = Moles of ethane / Total moles of mixture
Mole fraction of ethane = 0.93 moles / (0.93 moles + 0.70 moles)
Mole fraction of ethane ≈ 0.570
Mole fraction of ethene = Moles of ethene / Total moles of mixture
Mole fraction of ethene = 0.70 moles / (0.93 moles + 0.70 moles)
Mole fraction of ethene ≈ 0.430
Therefore, the composition of the original mixture was approximately 57% ethane and 43% ethene.