$$N_2 + 3H_2 \rightarrow 2NH_3$$

From the stoichiometry of the reaction, we can see that 1 mole of N2 requires 3 moles of H2 to produce 2 moles of NH3.

Therefore, for 14.5 moles of N2, we need:

$$14.5 \text{ mol }N_2 \times \frac{3 \text{ mol }H_2}{1 \text{ mol }N_2} = 43.5 \text{ mol }H_2$$

So, theoretically, 43.5 moles of H2 are needed to produce 22.5 moles of NH3 according to the given reaction.