Question

The first step in the Ostwald process for producing nitric acid is 

4NH₃(g) +5O₂(g) → 4NO(g) + 6H₂O(g). 

If the reaction of 0.15kg of ammonia with 0.15kg of oxygen gas yields 87g of nitric oxide, what is the percent yield of this reaction?

Answer 1

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 150.0 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(1.5*10^2 g)/(17.03 g/mol)

= 8.806 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 150.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(1.5*10^2 g)/(32 g/mol)

= 4.688 mol

Balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

4 mol of NH3 reacts with 5 mol of O2

for 8.806 mol of NH3, 11.01 mol of O2 is required

But we have 4.688 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of NO,

MM = 1*MM(N) + 1*MM(O)

= 1*14.01 + 1*16.0

= 30.01 g/mol

According to balanced equation

mol of NO formed = (4/5)* moles of O2

= (4/5)*4.688

= 3.75 mol

use:

mass of NO = number of mol * molar mass

= 3.75*30.01

= 1.125*10^2 g

% yield = actual mass*100/theoretical mass

= 87*100/1.125*10^2

= 77.31%

Answer: 77 %

Answer 2