This is a stoichiometry question, and will be answered in two steps, but the first thing for every stoichiometry question is to have the reaction properly balanced, which the question already provided us:
S8 + 8 O2 -> 8 SO2
First step is to find the limiting and excess reactant, this will tell us which element is in excess and by how much, let's gather the data for the reactants first:
30.14 grams of S8, molar mass of S8 is 256.5g/mol
287.43 grams of O2, molar mass of O2 is 32g/mol
Now we will focus on S8, and see how many moles we have of it in this reaction:
256.5 g = 1 mol
30.14 g = x moles
x = 0.12 moles of S8 in this reaction
According to the molar ratio, 1:8, we need 1 mol of S8 and 8 moles of O2 in order to have the reaction to proceed
1 S8 = 8 O2
0.12 S8 = x O2
x = 0.96 moles of O2 if we have 0.12 moles of S8
We only need 0.96, if we have more than that, is excess, if we have less than that, this means that O2 is the limiting and S8 is in excess, so let's check how much of O2 we do have
32g = 1 mol
287.43 g = x moles
x = 8.98 moles of O2, we have a big excess of O2, we only need 0.96 moles and we have almost 9 times more
Now that we know which element is the limiting, we can use this information to find the theoretical yield of SO2
The molar ratio will also be 1:8, and since we have 0.12 moles of S8, we will have 8 times that, which is 0.96 moles of SO2, using its molar mass, 64.1g/mol, we will have the theoretical yield
64.1g = 1 mol
x grams = 0.96
x = 61.5 grams is the theoretical yield
55.59 grams is the actual yield
Now we use these two values in the percent yield formula:
%yield = actual yield/theoretical yield
%yield = 55.59/61.5
%yield = 0.90
Limiting reactant = S8
Excess reactant = O2
Percent yield = 90%