Question

One of the steps in the manufacture of nitric acid is the oxidation of ammonia shown in this equation:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

If 47.0 kg of NH₃ reacts with 120. kg of O₂, what mass of NO is formed?

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)= 47.0 Kg = 47000 g

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(47000.0 g)/(17.034 g/mol)

= 2.759*10^3 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 120 Kg = 120000 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(120000.0 g)/(32 g/mol)

= 3.75*10^3 mol

Balanced chemical equation is:

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

4 mol of NH3 reacts with 5 mol of O2

for 2759.1875 mol of NH3, 3448.9844 mol of O2 is required

But we have 3750 mol of O2

so, NH3 is limiting reagent

we will use NH3 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 = (4/4)* moles of NH3

= (4/4)*2759.1875

= 2759 mol

mass of NO = number of mol * molar mass

= 2.759*10^3*30.01

= 8.28*10^4 g

Answer: 8.28*10^4 g