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Taking into account the reaction stoichiometry, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.
<h3>Reaction stoichiometry</h3>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: 14 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 ×14 g/mole= 14 grams
- H₂: 3 moles ×2 g/mole= 6 grams
- NH₃: 2 moles ×17 g/mole=34 grams
The following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 2 moles of NH₃, 16 grams of H₂ form how many moles of NH₃?
<u><em>moles of NH₃= 5.33 moles</em></u>
Then, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.
Learn more about the reaction stoichiometry:
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