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.

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)

What is the maximum mass of H2O that can be produced by combining 21.6 g of each reactant?

STRATEGY:

Convert 21.6 g NH3 to moles, then find the corresponding amount of H2O.

Convert 21.6 g O2 to moles, then find the corresponding amount of H2O.

Identify the limiting reactant and calculate the mass of H2O produced.

Step 1: 21.6 g NH3 and excess O2 produce 1.90 mol H2O.

Step 2: 21.6 g O2 and excess NH3 produce 0.810 mol H2O.

Step 3: When 21.6 g NH3 and 21.6 g O2 are mixed, how many moles of H2O form?

(1) 1.90 + 0.810 mol H2O

(2) 0.810 mol H2O

(3) 1.90 mol H2O

Calculate the mass of H2O produced from the number of moles of H2O.

mass: [???] g H2O

Answer 1

molar mass of NH₃ = 17.031 gm/mol then 21.6 gm of NH₃ = 21.6/17.031 = 1.268 mole

molar mass of O₂ = 32 g/mol then 21.6 gm O₂ = 21.6/32 = 0.675 mole

According to balanced chemical reaction to react with 4 mole of NH₃ requied O₂ = 5 mole therefore to react with 1.268 mole of NH₃ required mole of O₂ = 1.268 X 5 / 4 = 1.585 mole of O₂ but O₂ given only 0.675 mole therefore O₂ is limiting reactant.

According to balanced chemical reaction 5 mole of O₂ produce 6 mole of H₂O then 0.675 mole of O₂ produce

0.675 X 6 / 5 = 0.81 mole of H₂O

molar mass of H₂O = 18.01528 g/mol then 0.81 mole of H₂O = 0.81 X 18.01528 = 14.59 gm

14.59 gm of H₂O produced