if 2 litres of n2 is mixed with 2 litres of h2 calculate the volume of nh3 formed

Given : 
 Volume of N₂ = 2 L
 Volume of H₂ = 2 L
​To find : 
Volume of  NH₃ formed = ? 

Solution : 
           $N_2 + 3H_2 \to 2N{H}_3$
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$\times$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  $\frac{2}{3}$ volume of NH₃
​2 L of H₂ produce  $\frac{2\times 2}{3}$  L of NH₃ = 1.33 L of NH₃
Therefore, 1.33 L of NH₃ is formed.