Answer:
[tex]\large\boxed{\text{4 mol} }[/tex]
Explanation:
You haven't given the reaction. However, if all the N in NH₃ comes from N₂, we can write the partial equation
N₂ + … ⟶ 2NH₃ + …
n/mol: 2
2 mol of NH₃ are formed from 1 mol of N₂
[tex]\text{Moles of NH}_{3} = \text{2 mol N}_{2} \times \dfrac{\text{2 mol NH}_{3}}{\text{1 mol N}_{2}} = \textbf{4 mol NH}_{3}\\\\\text{The reaction produces $\large\boxed{\textbf{4 mol}}$ of NH}_{3}[/tex]
