Here, balanced equations :--
N2 + 3H2 ---> 2NH3

Since equal volumes of nitrogen and hydrogen are used
let's say 2 mole each...ok...

Nitrogen (2/1 = 2)
Hydrogen (2/3 = 0.66)

the lesser ratio is limiting reagent...

Here, Hydrogen is less...So, Hydrogen is limiting reagent...

$$Hope it's help... $$

Limiting Reagent in the Reaction of N2 and H2 to form Ammonia

Introduction:
In this reaction, N2 and H2 react to form Ammonia. The balanced chemical equation for the reaction is:

N2 + 3H2 → 2NH3

In this equation, 1 mole of Nitrogen (N2) reacts with 3 moles of Hydrogen (H2) to form 2 moles of Ammonia (NH3).

Limiting Reagent:
The limiting reagent is the reactant that gets consumed completely in a chemical reaction. It determines the amount of product that can be formed. The other reactants that are not consumed completely are called excess reagents.

In this reaction, we have equal volumes of N2 and H2. However, volumes cannot be used to determine the amount of substance in a chemical reaction. We need to use the molar ratio to determine the limiting reagent.

Molar Ratio:
The molar ratio is the ratio of the number of moles of one substance to the number of moles of another substance in a chemical equation. In this equation, the molar ratio of N2 to H2 is 1:3.

To determine the limiting reagent, we need to calculate the amount of each reactant in moles and compare it to the molar ratio.

Assuming we have 1 mole of N2 and 3 moles of H2, the amount of N2 and H2 in moles are:

N2 = 1 mole
H2 = 3 moles

Since the molar ratio of N2 to H2 is 1:3, we can see that we have enough H2 to react with the N2. Therefore, N2 is the limiting reagent.

Conclusion:
In the reaction of N2 and H2 to form Ammonia, the limiting reagent is Nitrogen (N2). It gets consumed completely in the reaction, and the amount of Ammonia formed is determined by the amount of N2 that reacts.