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 58.5 g of each reactant?

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)

mass:

g H2O

Answer 1

4 NH3 + 5 O2 = 4 NO + 6 H2O  

Moles of NH3 = mass / molar mass

NH3 molar mass = 17 grams/ mol

= 58.5 / 17

= 3.441

Molar mass of O2 = 32 grams / mol

Moles of O2 = 58.5 / 32

= 1.828

Now moles of O2 according to amount of NH3 in a reaction = 5/4 × moles of NH3 ( look balance equation)

= 5/4 × 3.441

= 4.30125

But we have only 1.828 moles of O2 , so O2 is limiting reagent. Moles of O2 will decide the amount of H2O

Now moles of H2O = 6/5 × moles of O2

= 6/5 × 1.828

= 2.1936

Mass of H2O = mole × molar mass

= 2.1936 × 18

= 39.4848 grams

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