Question

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of 6.0 grams of O2 with 7.0 grams of S. What is the % yield of SO3 in this experiment

Answer 1

Answer:

• 79%

Explanation:

1) Balanced chemical equation:

• 2S + 3O₂ → 2SO₃

2) Mole ratio:

• 2 mol S : 3 mol O₂ : 2 mol SO₃

3) Limiting reactant:

• Number of moles of O₂

        n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

• Number of moles of S:

         n = 7.0 g / 32.065 g/mol = 0.2183 mol S

• Ratios:

        Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

        Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

4) Calcuate theoretical yield (using the limiting reactant):

• 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

• x = 0.1875 × 2 / 3 mol SO₃ =  0.125 mol SO₃

5) Yield in grams:

• mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol =  10.0 g

6) Percent yield:

• Percent yield, % = (actual yield / theoretical yield) × 100

• % = (7.9 g / 10.0 g) × 100 = 79%