Q1. The equation for the reaction between sulfur dioxide and oxygen is shown.
2 SO2 (g) + O2 (g) = 2 SO³(g)

In an experiment, 2.00 mol of sulfur dioxide are mixed with 2.00 mol of oxygen.
The total amount of the three gases at equilibrium is 3.40 mol
What is the mole fraction of sulfur trioxide in the equilibrium mixture?
A 0.176

B 0.353

C 0.600
D 1.200

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

We have to use algebra here,

Because in the equation [tex]O_2[/tex] has a stoichometry of 1, the other have 2,

let [tex]O_2[/tex] = x

You started with 2 moles, so the moles of [tex]O_2[/tex] at equilibrium will be 2 - X

[tex]SO_3[/tex] has a stoichiometry of 2 compared to [tex]O_2[/tex], so will be 2X

( not minus from anything as you started with no moles of [tex]SO_3[/tex])

And then [tex]SO_2[/tex] will also be 2X as it’s stoichiometry is 2, but as you started with 2 moles it will be 2-2X

If you add these in an equation to get total moles it’s

2 - 2X + 2 - X + 2X = 3.4 (total given)

and then gathering together becomes 4-X = 3.4

so,

4 = 3.4 + X

so,

x = 0.6

Therefore,

the moles of [tex]SO_3[/tex] = 2 x 0.6 = 1.2

mole fraction = moles of the gas / total moles of gas

mole fraction = 1.2 / 3.4 = 0.353

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