For the reaction SO3 + H2O --> H2SO4, calculate the percent yield if 50g of water reacts with excess sulfur trioxide to produce 205g of hydrogen sulfate.

Answer 1

First, I see that the reaction is balanced as is. The stoichiometry is (50g H₂O) * (1 mol H₂O / 18.02g H₂O) * (1 mol H₂SO₄ / 1 mol H₂O) * (98.08g H₂SO₄ / 1 mol H₂SO₄) --> 272g H₂SO₄ theoretical.

Percent yield = 100 * actual/theoretical = 205/272 = 75.3%