Question

When 14.5 of SO2 reacts with 21 g of O2 , what will be the theoretical yield and percentage yield of the reaction if the actual yield is 12 g?

Answer 1

Answer:

Sorry I don't know that

Answer 2

The theoretical yield is 18.125 g and Percentage yield of the reaction is 66.207%

Find:

Theoretical yield and percentage yield of the reaction

Given:

Mass of SO₂ = 14.5 g

Mass of O₂ = 21 g

Actual yield = 12 g

Solution:

• First we should write the balanced chemical equation.
• 2SO₂ + O₂ ⇒2SO₃
• Molecular mass of SO₂ = 64 g
• There are 2 moles of SO₂, so total weight = 64*2 = 128 g
• 128 g of SO₂ reacts with 32 g of oxygen
• 14.5 g of SO₂ reacts with x g of oxygen
• X = $\frac{32*14.5}{128}$ = 3.625 g
• But we are provided with 21 g oxygen, so limiting reagent is SO₂.
• 2 moles of SO₂ gives 2 moles of SO₃
• 128 g SO₂ gives 160 g SO₃
• 14.5 g SO₂ = y g SO₃
• y = 18.125 g
• Theoretical yield = 18.125
• Actual yield = 12 g
• Percentage yield = actual yield/theoretical yield × 100
• = 12/18.125 × 100
• = 66.207 %
• So the theoretical yield is 18.125 g and Percentage yield of the reaction is 66.207%
• #SPJ2