At 1.00 atm and 0 oc

Question

At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 18.0 of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

Xmethane=

Xpropane=

Help I tried

Answer 1

So, this is a little more complicated than it appears to be at first glance. First, let us assume that the mass of CO₂ produced is 18.0 GRAMS, since you omitted the units (assume grams). Next, note that the conditions given (0ºC and 1 atm) represent STP.

First, write correctly balanced equations for both combustion reactions:

CH₄ + 2O₂ ==> CO₂ + 2H₂O

C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O

Next, since the conditions are at STP, we know that 1 mole of gas = 22.4 L, so we can find moles of gas present (CH4 + C3H8):

5.04 L x 1 mol / 22.4 L = 0.225 moles of gas

Now, let x = moles of CH₄ and then 0225-x = moles of C₃H₈. Calculate moles CO₂ from each reaction:

For CH₄: 1 mol CH₄ = 1 mol CO₂ so x mols CH₄ = x mols CO₂

For C₃H₈: 1 mol C₃H₈ = 3 mol CO₂ so 0.225-x mols C₃H₈ = 3* that = 0.675-3x mols CO₂

Next, we will calculate the mass (grams) of CO₂ produced in each reaction:

For CH₄: x mols CO₂ x 44 g / mol = 44x g CO₂

For C₃H₈: 0.675-3x mols CO₂ x 44 g / mol = 29.7 - 132x g CO₂

Since we know the mass of CO₂ produced is 18.0 g (given in the question), we can solve for x:

44x + 297 - 3x = 18.0

x = 0.133

And since x = moles CH₄, and 0.225-x = moles C₃H₈, we can now solve for moles of each gas:

moles CH₄ = 0.133 moles

moles C₃H₈ = 0.225 - 0.133 = 0.092 moles

X_methane = 0.133 / 0.225 = 0.591

X_propane = 0.092 / 0.225 = 0.409