Question

Hydrazine, N2H4, may react with oxygen to form nitrogen gas and water.

N2H4(aq)+O2(g)---------N2(g)+2H20(l)

If 3.05 g of N2H4 reacts and produces 0.850 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

First we have to calcuate the mass of N2.

Ideal gas equation , PV = n R T , R = 0.0821 L.atm/mol/K

Then, PV = (m/M) RT ( n = moles = mass (m)/ molar mass (M)

m = PVM/ RT

= ( 1 atm) (0.85 L) (28 g/mol) / (0.0821 L.atm/mol/K x 295 K )

= 0.983 g

Therefore,

mass of N2 produced = 0.983 g

          N2H4(aq) +   O2(g)   ---------> N2(g)+2H20(l)

            1 mol                                       1 mol

         32 g                                              28 g

         3.05 g                                          ?

         ? = ( 3.05 g/32 g ) x 28 g N2

             = 2.67 g of N2

   This theoretical yield of N2.

But, actual yield of N2 = 0.983 g

Therefore,

percent yield of the reaction = ( actual yield of N2/ theoretical yield of N2) X 100

                                                = ( 0.983 g / 2.76 g) X 100

                                                = 36.8 %