You have 28*g dinitrogen, and 24*g of dihydrogen. What is the limiting reagent with respect to ammonia formation?

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \rightarrow 2 N {H}_{3} \left(g\right)$
You have approx $1$ mol ${N}_{2}$ and $12$ mol ${H}_{2}$. If you assume 100% yield, 2 mol ammonia should result. Of course, the actual production of ammonia gets nowhere near these yields.