Result:
94.7 %
Explanation:
The balanced reaction is:
2 S + 3 O₂ → 2 SO₃
The stoichiometric mole ratio is:
S: 2 moles
O₂: 3 moles
Moles are calculated as mass divided by molar mass:
n = w / m
where n = moles, w = mass, m = molar mass.
Given:
For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol
For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol
Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.
Using proportions:
3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.
Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.
Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g
The percentage yield is actual yield divided by theoretical yield times 100:
Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %
