Stoichiometry: Limiting & Excess Reactants

If 5.0 mol of zinc sulfide reacts with 6.0 mol oxygen:
*which is the limiting reactant?
*which is the excess reactant?
*how many moles of sulfur dioxide will be made?

Reaction:
2 ZnS + 3 O2 --> 2 ZnO + 2 SO2

a walkthrough would be appreciated, thanks!

Okay to figure this out you have to find the mole ratio from the balanced equation of the two compounds you are comparing, in this case zinc and oxygen. The mole ratio is 2:3
Now make two sets of the mole ratio
1.)See how much oxygen can be produced with the given amount of zinc
2 ZnS: 3 O2
5 mol(given): x
solve for x: (5X3)/2 = 7.5 mol of O2 produced

2.) See how much zinc can be produced with the given amount of oxygen
2 ZnS: 3 O2
x: 6 mol(given)
solve for x: (6X2)/3 = 4 mol of ZnS produced

Okay now compare the amount produced with the amount provided.
There's 5 mol of ZnS and you can produce 4 mol. This means there is enough
There's 6 mol O2 and you can produce 7.5 mol. This means that there is more given than can be produced and therefore O2 is the limiting reactant and ZnS is the excess reactant!

you'll have to use stoichiometric ratios
if you have 5 mol of ZnS, and there is a 2:2 ratio of ZnS to SO2 (from balanced formula), that means there could theoretically be 5 mol of SO2 (5 mol ZnS) * (2 mol SO2 / 2 mol ZnS) = 5 mol SO2 --> if you write that, you'll see the mol of ZnS will cancel out because it will be in both the numerator and denominator

if you have 6 mol O2, and there is a 3:2 ratio of O2 to SO2....so (6 mol O2) * (2 mol SO2 / 3 mol O2)=4 mol SO2

since the O2 yields less SO2, it is the limiting reactant; since the ZnS yields more SO2, it is the excess reactant

4 moles of SO2 will be made