Why did the chemist break up with his SO2(g) molecule?
Because it wasn't giving him enough SO3(g) in return!
But let's get serious for a moment and do some calculations.
We're given that the equilibrium constant (K) for this reaction is 2.00. At equilibrium, the ratio of products to reactants can be determined using the stoichiometry of the reaction.
From the balanced equation, we can see that for every 1 mole of SO2(g) that reacts, 1 mole of SO3(g) is produced. And for every 1 mole of SO2(g) that reacts, 1 mole of NO2(g) reacts as well.
So, if we start with 2.64 mol of SO2(g) and want to form 1.10 mol of SO3(g), we can determine the required amount of NO2(g) using the stoichiometry.
Since the ratio of SO2(g) to SO3(g) is 1:1, we can say that if we form 1.10 mol of SO3(g), we need 1.10 mol of SO2(g). And since the ratio of SO2(g) to NO2(g) is also 1:1, we would need 1.10 mol of NO2(g) as well.
Therefore, you would need to add 1.10 mol of NO2(g) to 2.64 mol of SO2(g) in order to form 1.10 mol of SO3(g) at equilibrium.