Question

In the Raschig Process, ammonia, NH3, is oxidized with sodium hypochlorite (formed from NaOH and Cl2) to produce hydrazine.

a) If 25.0 grams of NaOH reacted with 30.00 grams of Cl2, in the presence of excess NH3, which is the limiting reactant, and what is the theoretical yield of N2H4?

b) If 8.00 grams of N2H4 is recovered, what is the percent yield?

Answer 1

Solution - first see balance reaction of formation of hydrazine -

2NaOH + Cl₂ + 2NH₃$\LARGE \rightarrow$ N₂H₄ + 2NaCl + 2H₂O

a)  

From balance reaction it is clear that one mole of chlorine gas required two moles of NaOH , were NH₃ is in excess, so we have to check moles of NaOH and chlorine only.

# Molar mass of NaOH = 40 g/mol

# Molar mass of chlorine gas = 70.9 g/mol

#Molar mass of hydrazine = 32.04 g/mol

#mass of NaOH = 25 .0 g

#mass of chlorine gas = 30.0 g

Now we can calculate number of moles by using molar mass and mass of individual component.

Number of moles (n) = (mass of NaOH /Molar mass of NaOH)

= (25.0 g /40.0 g / mol )

=0.625 mol of NaOH

Now we have to calculate number of moles of chlorine gas =(mass of chlorine gas /molar mass of chlorine gas)

=(30.0 g /70.9 g/mol)

= 0.4231 mol of chlorine gas

But from balance reaction it is clear that one mole of chlorine gas required two moles of NaOH, so theoretically moles require for complete reaction = (moles of chlorine gas x 2)

= (0.4231 mol x 2)

= 0.8462 mol of NaOH required but we have only 0.625 mol, so limiting reagent is NaOH ( 0.8462 mol (theoretical moles)  $\LARGE >$ 0.625 mol)

Now we have to calculate theoretical yield of hydrazine against limiting moles of NaOH ,

Theoretical yield of hydrazine = ( moles of NaOH x molar mass of hydrazine)

= (0.625 mol x 32.04 g/mol )

= 20.0 g

b)

Amount of hydrazine recovered ( practical yield) = 8.00 g, what is percentage yield?

We know that,

percentage yield = (practical yield /Theoretical yield) x 100

= (8.00 /20.0 )x100

= 40%