Answer:
Percent yield = 57.2%
Explanation:
Given data;
Mass of nitrogen = 15.0 g
Mass of hydrogen = 15.0 g
Mass of ammonia produced = 10.5 g
Percent yield = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 15 g / 28 g/mol
Number of moles = 0.54 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 15 g / 2 g/mol
Number of moles = 7.5 mol
Now we compare the moles ammonia with hydrogen and nitrogen
N₂ : NH₃
1 : 2
0.54 : 2/1 ×0.54 = 1.08 mol
H₂ : NH₃
3 : 2
7.5 : 2/3×7.5= 5 mol
The number of moles of ammonia produced by nitrogen are less so it will limiting reactant.
Theoretical yield:
Mass = number of moles × molar mass
Mass = 1.08 mol × 17 g/mol
Mass = 18.36 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 10.5 g / 18.36 g × 100
Percent yield = 57.2%
