Answer:
0.0116 mol.
Step-by-step explanation:
- Suppose that the reaction proceeds at STP conditions.
It is known that at STP: 1 mol of any gas occupies 22.4 L.
Using cross multiplication:
1 mol of NH₃ occupies → 22.4 L.
??? mol of NH₃ occupies → (520 mL) 0.52 L.
∴ The no. of moles of (520 mL) NH₃ produced = (1 mol)(0.52 L)/(22.4 L) = 0.0232 mol.
- For the balanced reaction:
3H₂ + N₂ → 2NH₃.
3 mol of H₂ react with 1 mol of N₂ to produce 2 mol of NH₃.
Using cross multiplication:
1 mol of N₂ produce → 2 mol of NH₃, from stichiometry.
??? mol of N₂ produce → 0.0232 mol of NH₃.
∴ The no. of moles of N₂ needed = (1 mol)(0.0232 mol)/(2 mol) = 0.0116 mol.