Question

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

N2H4(aq)+O2(g)⟶N2(g)+2H2O(l)

If 3.15 g of N₂H₄ reacts and produces 0.750 L of N₂, at 295 K and 1.00 atm, what is the percent yield of the reaction?

%

Answer 1

Step 1: Calculate theoretical yield in mol

Molar mass of N2H4,

MM = 2*MM(N) + 4*MM(H)

= 2*14.01 + 4*1.008

= 32.052 g/mol

mass of N2H4 = 3.15 g

mol of N2H4 = (mass)/(molar mass)

= 3.15/32.05

= 9.828*10^-2 mol

According to balanced equation

mol of N2 formed = moles of N2H4

= 9.828*10^-2 mol

step 2: calculate actual yield in mol

Given:

P = 1.0 atm

V = 0.75 L

T = 295.0 K

find number of moles using:

P * V = n*R*T

1 atm * 0.75 L = n * 0.08206 atm.L/mol.K * 295 K

n = 3.098*10^-2 mol

step 3: calculate percent yield

% yield = actual * 100 / theoretical

= (3.098*10^-2)*100 / (9.828*10^-2)

= 31.5 %

Answer: 31.5 %