The reaction of N2 and H2 to generate NH3 is as follows:N2 + 3H2 ⇒2NH3
Since H2 is in excess amount in this reaction, the determine factor is N2. The ratio of N2 and product NH3 is 1:2. Therefore, 4.08 moles of N2 can generate 4.08*2 = 8.16 mole of NH3.
Grams of NH3 = moles of NH3 * molar mass of NH3 = 8.16 mole * 17 g/mole = 139 g.
139 grams of nh3 can be produced from 4.08 mol of n2 and excess h2.
