Answer:
1) The limiting reactant is Nâ‚‚ because it is present with the lower no. of moles than Hâ‚‚.
2) The amount (in grams) of excess reactant Hâ‚‚ = 4.39 g.
Explanation:
- Firstly, we should write the balanced equation of the reaction:
N₂ + 3H₂ → 2NH₃.
1) To determine the limiting reactant of the reaction:
- From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- This means that Nâ‚‚ reacts with Hâ‚‚ with a ratio of (1:3).
- We need to calculate the no. of moles (n) of Nâ‚‚ (5.23 g) and Hâ‚‚ (5.52 g) using the relation: n = mass / molar mass.
The no. of moles of Nâ‚‚ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.
The no. of moles of Hâ‚‚ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.
- From the stichiometry, Nâ‚‚ reacts with Hâ‚‚ with a ratio of (1:3).
The ratio of the reactants of Nâ‚‚ (5.23 g, 0.1868 mol) to Hâ‚‚ (5.52 g, 2.74 mol) is (1:14.67).
∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
0.1868 mol of Nâ‚‚ react completely with 0.5604 mol of Hâ‚‚ and the remaining of Hâ‚‚ is in excess.
2) To determine the amount (in grams) of excess reactant of the reaction:
- As showed in the part 1, The limiting reactant is Nâ‚‚ because it is present with the lower no. of moles than Hâ‚‚.
- Also, 0.1868 mol of Nâ‚‚ react completely with 0.5604 mol of Hâ‚‚ and the remaining of Hâ‚‚ is in excess.
- The no. of moles are in excess of Hâ‚‚ = 2.74 mol - 0.5604 mol (reacted with Nâ‚‚) = 2.1796 mol.
- ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.
