In this problem, we can solve this by comparing which of the reactants will produce the least amount of product.

`2 SO_2 (g) + O_2 (g) -> 2 SO_3 (g)`

So we will solve for the number of moles of `SO_3` produced using the two moles given.

Using `SO_2`

`3.1 mol es SO_2 * (2 mol es SO_3)/(2 mol es SO_2) `

= 3.1 moles `SO_3`

Using `O_2`

`2.7 mol es O_2 * (2 mol es SO_3)/(1 mol es O_2) `

= 5.4 moles `SO_3`

We can see that the limiting reagent is `SO_2` and the excess reagent is the `O_2` and therefore we will use the amount of `SO_3` derived from the moles of `SO_2` . 

From `SO_2` :

 `3.1 mol es SO_3 *(80.0632 grams)/(1 mol e SO_3)` ` `

= 248.1959 = 248 grams `SO_3`