Question

a) suppose a chemist mixes 13.8 grams of ethane and 45.8 grams of dioxygen. Calculate the theoretical yeild of water.

b) Suppose the reaction actually produces 14.2 grams of water. Calculate the percent yeild of water.

Answer 1

   2H2 + O2 -------------> 2H2O

no of moles of H2 = W/G.M.Wt

                                = 13.8/2    = 6.9 moles

no of moles of O2 = W/G.M.Wt

                               = 45.8/32   = 1.43moles

from balanced equation

2 moles of H2 react with 1 mole of O2

6.9 moles of H2 react with = 6.9*1/2    = 3.45 moles of O2 is required.

O2 is limiting reagent

1 mole of O2 react with H2 to gives 2 moles of H2O

1.43 moles of O2 react with H2 to gives = 2*1.43/1    = 2.86 moles of H2O

mass of H2O = no of moles * gram molar mass

                       = 2.86*18   = 51.48g >>>>>theoretical yield

percentage yield = actual yield*100/theoretical yield

                             = 14.2*100/51.48   = 27.6% >>>>>>answer