Question

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H2O that can be produced by combining 60.4 g of each reactant?

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)= 60.4 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(60.4 g)/(17.03 g/mol)

= 3.546 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 60.4 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(60.4 g)/(32 g/mol)

= 1.887 mol

Balanced chemical equation is:

4 NH3 + 5 O2 —> 4 NO + 6 H2O

4 mol of NH3 reacts with 5 mol of O2

for 3.546 mol of NH3, 4.432 mol of O2 is required

But we have 1.887 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

According to balanced equation

mol of H2O formed = (6/5)* moles of O2

= (6/5)*1.887

= 2.265 mol

use:

mass of H2O = number of mol * molar mass

= 2.265*18.02

= 40.81 g

Answer: 40.8