N2(g) + 3H2(g) → 2NH3(g)
\(\because\) 1 L of N2 reacts with 3 L of H2.
\(\therefore\) 2 L of N2 will react with 6 L of H2, but we have only 2 L of H2, So, H2 is the limiting reagent.
\(\because\) 3 L of H2 gives 2 L of NH3
\(\therefore\) 2 L of H2 will give = \(\frac{2}{3}\) x 2 = \(\frac{4}{3}\) = 1.33 L of NH3