Question

1) Beginning with 1.0 g of trans-stilbene, 2.0 mL of concentrated aqueous HBr (45% w/w) and 1.6 mL of 30% w/w hydrogen peroxide, identify the limiting reagent and calculate the theoretical yield of the 1,2 dibromoalkane. If you isolated 0.95 g of 1,2-dibromo-1,2-diphenylethane, what would be your percent yield?

Answer 1

Solution :-

1g Stilbene

2ml 45 % w/w HBr

So mass of HBr = 45 * 2/100 = 0.9 g HBr

1.6ml of 30 % H2O2

So mass of H2O2 = 30 % * 1.6 / 100 = 0.48 g H2O2

Now lets write the balanced reaction equation

C14H12 + 2HBr + H2O2 ------ > C14H12Br2 + 2H2O

Now lets calculate the moles of each reactant

Moles of stilbene C14H12 = 1 g / 180.25 g per mol = 0.005548 mol

Moles of HBr = 0.9 g / 80.9119 g per mol = 0.01112 mol

Moles of H2O2 = 0.48 g / 34.02 g per mol = 0.01411 mol H2O2

Lets find the moles of the HBr needed to react with stilbene

0.005548 mol stilbene * 2 mol HBr / 1 mol stilbene = 0.01110 mol

So the moles of HBr are enough to react with stilbene

The mole ratio of the stilbene and H2o2 is 1 :1

So the moles of H2O2 are also enough to react with stilbene

So the limiting reactant is stilbene

Now lets calculate the moles of product using the moles of stilbene

Moles of product C14H12Br2 = 0.005548 mol stilbene * 1 mol product / 1 mol stilbene

                                                         = 0.005548 mol product

Now lets calculate the mass of the product

Mass of the product C14H12Br2 = moles * molar mass

                                                           =0.005548 mol * 340.053 g per mol

                                                      = 1.886 g product

Theroetical yield = 1.886 g

Now lets calculate the percent yield

% yield = (actuall yield / theoretical yield )*100%

              = (0.95 g / 1.886 g)*100%

               = 50.37 %