Taking into account the reaction stoichiometry, 56.42 grams of N₂ are formed when 4 moles of NH₃ react.
Reaction stoichiometry
In first place, the balanced reaction is:
4 NH₃ + 3 O₂ → 2 N₂+ 6 H₂O
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:
- NH₃: 4 moles
- O₂: 3 moles
- N₂: 2 moles
- H₂O: 6 moles
The molar mass of the compounds is:
- NH₃: 17 g/mole
- O₂: 32 g/mole
- N₂: 28 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- NH₃: 4 moles ×17 g/mole= 68 grams
- O₂: 3 moles ×32 g/mole= 96 grams
- N₂: 2 moles ×28 g/mole= 56 grams
- H₂O: 6 moles ×18 g/mole= 108 grams
Mass of N₂ formed
The following rule of three can be applied: if by reaction stoichiometry 4 moles of NH₃ form 56 grams of N₂, 4.03 moles of NH₃ form how much mass of N₂?
[tex]mass of N_{2} =\frac{4.03 moles NH_{3} x56 grams of N_{2} }{4 moles NH_{3}}[/tex]
mass of N₂= 56.42 grams
Then, 56.42 grams of N₂ are formed when 4 moles of NH₃ react.
Learn more about the reaction stoichiometry:
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