What is the percent yield for this reaction under the given conditions? I think I have the right answer(16.6%), but can someone check it for me? thanks!

Answer:
Percentage yield = 16.4%
which almost equal to 16.6%. So the given answer is true.
Explanation:
Given data:
Mass of hydrogen = 1.52 g
Mass of nitrogen = 9.89 g
Mass of ammonia produced (Actual yield) = 1.42 g
Percentage yield = ?
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 1.52 g/ 2 g/mol
Number of moles = 0.76 mol
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 9.89 g/ 28 g/mol
Number of moles = 0.35 mol
Now we will compare the moles of NH₃ with nitrogen and hydrogen.
H₂ : NH₃
3 : 2
0.76 : 2/3 ×0.76 = 0.51 mol
N₂ : NH₃
1 : 2
0.35 : 2×0.35 = 0.7 mol
The number of moles of NH₃ produced from hydrogen are less that's why Hydrogen will be limiting reactant.
Theoretical yield:
Mass of NH₃ = moles × molar mass
Mass of NH₃ = 0.51 mol × 17 g/mol
Mass of NH₃ = 8.67 g
Theoretical yield of NH₃ = 8.67 g.
Percentage yield:
Percentage yield = actual yield / theoretical yield × 100
Percentage yield = 1.42 / 8.67 × 100
Percentage yield = 16.4%
which almost equal to 16.6%. So the given answer is true.