Question

When 20g of sulphur dioxide reacts with oxygen, 23g of sulphur trioxide is formed what is the percentage yields?

Answer 1

Answer:

Percent yield = 92 %

Explanation:

SO₂ + O₂ --------> SO₃

20g                     23g

1 mole of SO₂ = 1 mole of SO₃

SO₂ molar mass = 32 +2x16 = 64g

SO₃ molar mass = 32 + 3x16 = 80g

64g of SO₂ = 80 g of SO₃

20 g of SO₂ = x g of SO₃

x = 80 x 20 / 64

x = 25 g

Actual yield of SO₃ = 23g

Theoretical yield of SO₃ =  25 g

percent yield = Actual yield        x 100

                         Theoretical yield

                   = 23 x 100 / 25

                   = 92 %

Thank You !!