when 1 mole of n2 and 1 mole of h2
When 1 mole of N₂ and 1 mole of H₂ are allowed to react to form ammonia, H₂ is the limiting reagent and only a small amount of NH₃ can form.
Step 1: Use the balanced equation
The balanced equation for ammonia synthesis is:
N2+3H2→2NH3\text{N}_2+3\text{H}_2\rightarrow 2\text{NH}_3N2+3H2→2NH3
This shows that 1 mole of N₂ needs 3 moles of H₂ to react completely.
Step 2: Compare required vs available moles
- You have:
- N₂ = 1 mole
- H₂ = 1 mole
- Stoichiometric requirement:
- For 1 mole N₂ → need 3 moles H₂
- But only 1 mole H₂ is present, so H₂ is limiting and N₂ is in excess.
Step 3: Calculate moles of NH₃ formed
From the balanced equation:
- 3 moles H₂ → 2 moles NH₃
So for 1 mole H₂:
NH3=23 mole\text{NH}_3=\frac{2}{3}\text{ mole}NH3=32 mole
Therefore, when 1 mole of N₂ and 1 mole of H₂ react:
- Maximum NH₃ formed = 23\frac{2}{3}32 mole
- H₂ is completely consumed, and some N₂ is left unreacted.
So, the key result: 2 3\frac{2}{3}32 mole of NH₃ can be produced from 1 mole N₂ and 1 mole H₂.