Determine the mass of ammonia NH3 produced when 3.75 g of nitrogen gas N2 react with hydrogen gas H2

N2(g) + 3H2-> 2NH3(g)  This is the balanced equation

Note the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important.

moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present

moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present

Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.

moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced 

grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced

NOTE:  The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

(5.00g H2) * (1 mol H2 /2.016g H2) * (2 mol NH3 /3 mol H2) * (17.02g NH3 /1 mol NH3) = 28.3g NH3.