Answer:
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 %