Question

find the percent yield of product if 1.50 gram of SO2 is produced from 1.00 gram and excess sulfur via the reaction

Answer 1

   S + O2 ------------------> SO2

1 mole of O2 react with S to gives 1 mole of SO2

32g of O2 react with S to gives 64g of SO2

1g of O2 react with S to gives = 64*1/32   = 2 g of SO2

percentage yield = Actual yield*100/ theoretical yield

                           = 1.5*100/2   = 75% >>>>>answer