For the reaction represented by the equation 1SO3 1H2O -> 1H2SO4, calculate the percentage yield if 500. g of sulfur trioxide react with excess water to produce 575 g of sulfuric acid.

The answer to the question is

The percentage yield when 500. g of SO₃  react with excess water to produce 575 g of H₂SO₄ is 81.63 %

Explanation:

The known variables are

Mass of SO₃ = 500 g

Mass of H₂SO₄ = 575 g

Molar mass of SO₃ = 80.0632 g/mol

Molar mass of H₂SO₄ = 98.079 g/mol

The reaction equation is

SO₃ +H₂O → H₂SO₄

One mole of SO₃  react with one mole of H₂O to produce one mole of  H₂SO₄

Therefore number of moles of SO₃ in 500 g sample of SO₃ = (500 g)/(80.0632 g/mol) = 6.25 moles

Also number of moles of H₂SO₄ in 575 g sample of H₂SO₄ = (575 g)/(98.079 g/mol) = 5.86 moles

Which shows that the reaction is in the forward direction, production of H₂SO₄

Therefore 6.25 moles of SO₃ can produce 6.25 moles of H₂SO₄ and

The mass of 6.25 moles of H₂SO₄  = 6.25 moles × 98.079 g/mol = 612.51 g

Theoretical Yield = 612.51 g

Percentage Yield = =

Percentage Yield = 81.63 %