Ineed finding the percent yield for the following question, i am having a hard time understanding it.
find the percent yield of product if 1.50 g of so3 is produced from 1.00 g of o2 and excess sulfur via the reaction:
2s + 3o2 = 2so3

The formula for percent yield is (actual yield / theoretical yield)(100)
Your actual yield is how much of the product that was produced by doing the experiment. The theoretical yield is how much product should have been made given the circumstances.

In this reaction, you are told that 1.50g of SO3 is produced; this is your actual yield. 
To find the theoretical yield, you would have to do stoichiometry using the 1.00g of O2 that you are given. Since we are told that excess sulfur was used, we do not have to worry about finding the limiting reactant.
So: 
(1.00g O2)/(32.0g/mol O2)= 0.03125mol O2
(0.03125mol O2)(2mol SO3/3mol O2) = 0.021mol SO3
(0.021mol SO3)(80.1 g/mol SO3) = 1.67g SO3

1.67g is your theoretical yield.

Now using the formula for percent yield:
(1.50g SO3)/(1.67g SO3) = 0.898 x 100 = 89.8%, this is your percent yield :)