Question

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

N2H4(aq) + O2 (g) = N2 (g) + 2H2O (l)

If 3.25 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

The balanced equation is as follows:

N2H4(aq) + O2 (g) = N2 (g) + 2H2O (l)

1mol N2H4 will produce 1mol N2

Now calculate the moles of N2H4 in 3.25 g

Molar mass of N2H4 = 32.0452 g/mol
3.25 g N2H4 = 3.25/32.0452

= 0.10 mol N2

Now calculate the mol of N2 as follows:

1 mol of N2/ 1 mol N2H4* 0.10 mol N2

= 0.10 mol N2

Now calculate the moles of N2 of 0.850 L of N2, at 295 K and 1.00 atm

PV = nRT

n=PV/ RT/

n=1.00 atm* 0.850 l/ 0.08206 atm –L/ mol-K*295 K

n= 0.035 mol

% yield = 0.035/0.10*100

= 35.0 % yield.