If 2 L of N2 is mixed with 2 L of H2 at a constant temperature and pre

1.	If 2 L of N2 is mixed with 2 L of H2 at a constant temperature and pressure, then what will be the volume of NH3 formed?
Answer» 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

If 2 L of N2 is mixed with 2 L of H2 at a constant temperature and pressure, then what will be the volume of NH3 formed?

Answer»

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