N2 + 3H2 →2NH3

From the reaction it can be seen clearly that 1 volume of N2 is required to react with 3 volume of H​2 to give 2 volumes of NH3

1 volume of N2 react with  3 volume of H2

2 L of N2 react with  3×2 L of H2 = 6 L of H2

But we only have 2 L of H2. 
Therefore, H2 is the limiting reagent. 

3 volumes of H2 produce  2 volumes of NH3

1 volume of H2 produce  2/3 volume of NH3
​2 L of H2 produce  2×2/3  L of NH3 = 1.33 L of NH3
Therefore, 1.33 L of NH3 is formed.

Calculation of Volume of NH3 formed

To calculate the volume of NH3 formed, we need to follow the balanced chemical equation for the reaction between N2 and H2 to form NH3. The balanced chemical equation is as follows:

N2 + 3H2 → 2NH3

According to the equation, one mole of N2 reacts with three moles of H2 to form two moles of NH3. Therefore, we need to calculate the number of moles of N2 and H2 present in the mixture of 2 litres each.

Calculation of Moles of N2 and H2

The volume of the gas is directly proportional to the number of moles of gas present. The molar volume of any gas at standard temperature and pressure (STP) is 22.4 L/mol. Therefore, the number of moles of N2 and H2 present in the mixture can be calculated as follows:

Number of moles of N2 = Volume of N2 / Molar volume of N2 at STP
= 2 L / 22.4 L/mol
= 0.089 moles

Number of moles of H2 = Volume of H2 / Molar volume of H2 at STP
= 2 L / 22.4 L/mol
= 0.089 moles

Calculation of Limiting Reagent

To find out the limiting reagent, we need to calculate the number of moles of NH3 formed by each of the reactants. The reactant that forms the least amount of NH3 will be the limiting reagent.

Number of moles of NH3 formed by N2 = Number of moles of N2 x (2/1)
= 0.089 x (2/1)
= 0.178 moles

Number of moles of NH3 formed by H2 = Number of moles of H2 x (2/3)
= 0.089 x (2/3)
= 0.059 moles

Therefore, H2 is the limiting reagent as it forms the least amount of NH3.

Calculation of Volume of NH3 formed

The number of moles of NH3 formed by the reaction is equal to the number of moles of limiting reagent used. Therefore, the volume of NH3 formed can be calculated as follows:

Volume of NH3 formed = Number of moles of NH3 x Molar volume of NH3 at STP
= 0.059 x 22.4 L/mol
= 1.32 L

Therefore, 1.32 litres of NH3 is formed when 2 litres of N2 is mixed with 2 litres of H2 at constant temperature and pressure.