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Taking into account the reaction stoichiometry, 159.28 grams of NH₃ are formed from 4.42 moles of N₂ and excess H₂.

Reaction stoichiometry

In first place, the balanced reaction is:

N₂ + 3 H₂  → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • N₂: 1 mole
  • H₂: 3 moles
  • NH₃: 2 moles

The molar mass of the compounds is:

  • N₂: 28 g/mole
  • H₂: 2 g/mole
  • NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • N₂: 1 mole ×28 g/mole= 28 grams
  • H₂: 3 moles ×2 g/mole= 6 grams
  • NH₃: 2 moles ×17 g/mole= 34 grams

Mass of NH₃ formed

The following rule of three can be applied: if by reaction stoichiometry 1 mole of N₂ form 34 grams of NH₃, 4.42 moles of N₂ form how much mass of NH₃?

[tex]mass of NH_{3} =\frac{4.42 moles of N_{2}x34 grams of NH_{3} }{1 mole of N_{2}}[/tex]

mass of NH₃= 159.28 grams

Then, 159.28 grams of NH₃ are formed from 4.42 moles of N₂ and excess H₂.

Learn more about the reaction stoichiometry:

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