Question

You are provided with 6.05 g S O 2 to react with 2.23 g O 2 . Indicate your theoretical yield.

Answer 1

Write and balance the equation

2SO2 + O2 -> 2SO3

4SO2 + O2 -> 2S2O5

I assume is the first reaction

Calculate the theoretical yield for each of the reactants and the lowest yield will be the theoretical yield of the reaction produced by the limiting regent. Calculate the molecular mass of SO2 , SO3, and O2 from the atomic mass of the elements in the periodic table.

6.05 g SO2 x 80.1g SO3 / 64.1 g SO2 = 7.56 g SO3

2.23g O2 x 2* 80.1 g SO3 /32gO2 = 11.16 g SO3

Answer 2

7.56 g

Step-by-step explanation:

Data: Molar masses

SO₂: 64.07 g/mol

O₂: 32.00 g/mol

SO₃: 80.07 g/mol

Let's consider the following balanced reaction.

SO₂ + 1/2 O₂ → SO₃

The theoretical mass ratio of SO₂ to O₂ is 64.07g to 1/2 × 32.00 g = 16.00 g, i.e. 4.004 g SO₂/ 1 g O₂.

The experimental mass ratio of SO₂ to O₂ is 6.05 g to 2.23 g, i.e. 2.713 g SO₂/ 1 g O₂.

Comparing both ratios, we can determine that SO₂ is the limiting reactant.

The mass ratio of SO₂ to SO₃ is 64.07 g to 80.07 g. The theoretical yield of SO₃ from 6.05 g of SO₂ is:

6.05 g SO₂ × (80.07 g SO₃/ 64.07 g SO₂) = 7.56 g SO₃