The equation for the reaction between sulfur dioxide and oxygen is shown - AQA - A-Level Chemistry - Question 11 - 2019 - Paper 3

Now we write the total in terms of x:

$(2.00 - x) + (2.00 - \frac{x}{2}) + x = 3.40$

Simplifying this:

$4.00 - x + x - \frac{x}{2} = 3.40$

This simplifies to:

$4.00 - \frac{x}{2} = 3.40$

Isolating x gives:

$\frac{x}{2} = 0.60 \implies x = 1.20$

From our calculation, the moles of sulfur trioxide at equilibrium = 1.20 mol.

Now we can find the mole fraction of SO₃:

$\text{Mole fraction of } SO₃ = \frac{\text{Moles of } SO₃}{\text{Total moles}} = \frac{1.20}{3.40}$

Calculating this gives:

$\text{Mole fraction of } SO₃ = \frac{1.20}{3.40} \approx 0.353$

Thus, the mole fraction of sulfur trioxide in the equilibrium mixture is approximately 0.353.