Let's begin by writing a correctly balanced equation for the reaction taking place.
N2(g) + 3H2(g) ===> 2NH3(g)
Next, we will find the limiting reactant. There are various ways to accomplish this, but to me, one of the easiest is to calculate moles of each reactant, and then divide that number by the coefficient in the balanced equation. So, we have the following:
moles N2 = 100.0 g x 1 mol/28 g = 3.571 moles N2 (divided by 1 = 3.57)- This is less than 16.7 so is LIMITING
moles H2 = 100.0 g x 1 mol/2 g = 50 moles H2 (divided by 3 = 16.7) - Excess reactant
Once we have identified the limiting reactant, we use that to determine theoretical yield:
3.571 moles N2 x 2 moles NH3 / 1 mole N2 x 17 g NH3/mol NH3 = 121.4 g NH3