Answer:
13.82 %.
Explanation:
- The % yield is calculated from the relation:
% yield = [(actual mass of NH₃) / (theoretical mass of NH₃)] x 100.
The actual mass of NH₃ = 45.5 g.
- Now, we should calculate the theoretical mass of NH₃ produced upon reaction of 13.0 moles N₂ and 29.0 moles H₂.
- From the balanced equation (N₂ + 3H₂ → 2NH₃), it is clear that every 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- The limiting reactant is H₂ (29.0 moles) that requires 9.667 moles of N₂ (excess) to be completely reacted.
Using cross multiplication:
3.0 moles of H₂ produces → 2.0 moles of NH₃, from the stichiometry,
29.0 moles of H₂ produces → ??? moles of NH₃.
∴ The number of moles of NH₃ produced upon reaction of (13.0 moles N₂ and 29.0 moles H₂) = (2.0)(29.0) / (3.0) = 19.333 mol.
- Now, we can calculate the theoretical mass of NH₃ produced = n x molar mass = (19.333 mole)(17.031 g/mol) = 329.266 g.
∴ % yield = [(actual mass of NH₃) / (theoretical mass of NH₃)] x 100 = [(45.5 g) / (329.266 g)] x 100 = 13.82 %.
