In the production of ammonia, n2 + 3h2 --> 2nh3, if you start with 3 mol of n2 and 6 mol of h2, identify the limiting reagent and the excess reagent.

From the given chemical reaction, we determine the amount of NH3 formed from the amounts of the reactants.
 
      N2:     (3 mol N2) x (2 mols NH3 / 1 mole N2)  = 6 moles NH3
      H2:     (6 mol H2) x (2 mols NH3 / 3 moles H2)  = 4 moles NH3
Since there are fewer of the NH3 that can be formed from the H2. Then, it is the limiting reagent and the N2 is the excess reagent. 

60.66% is the answer