Answer:
Percent yield = 196%
Explanation:
Given data;
Mass of SO₂ = 364 g
Mass of O₂ = 42.0 g
Mass of SO₃ = 408 g
Percent yield = ?
Solution:
Chemical equation:
2SO₂ + O₂ → 2SO₃
Number of moles of SO₂:
Number of moles = mass/ molar mass
Number of moles = 364 g / 64.066 g/mol
Number of moles = 5.7 mol
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 42.0 g / 32 g/mol
Number of moles = 1.3 mol
Now we compare the moles ammonia with hydrogen and nitrogen
O₂ : SO₃
1 : 2
1.3 : 2/1 ×1.3 = 2.6 mol
SO₂ : SO₃
2 : 2
5.7 : 5.7
The number of moles of SO₃ produced by oxygen are less so it will limiting reactant.
Theoretical yield:
Mass = number of moles × molar mass
Mass = 2.6 mol × 80.07 g/mol
Mass = 208.2 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 408 g / 208.2 g × 100
Percent yield = 1.96 × 100
Percent yield = 196%
