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 62.6 g of each reactant?

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

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)= 62.6 g

we have below equation to be used:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(62.6 g)/(17.034 g/mol)

= 3.675 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 62.6 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(62.6 g)/(32 g/mol)

= 1.956 mol

we have the Balanced chemical equation as:

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

4 mol of NH3 reacts with 5 mol of O2

for 3.675 mol of NH3, 4.5938 mol of O2 is required

But we have 1.9563 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

From balanced chemical reaction, we see that

when 5 mol of O2 reacts, 6 mol of H2O is formed

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

= (6/5)*1.9563

= 2.348 mol

we have below equation to be used:

mass of H2O = number of mol * molar mass

= 2.348*18.02

= 42.3 g

Answer: 42.3 g