Taking into account the reaction stoichiometry, 3.63 moles of NH₃ will be produced if 50.8 g N₂ reacts with 187.3 g H₂.
In first place, the balanced reaction is:
3 H₂ + N₂ → 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:
The molar mass of the compounds is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 6 grams of H₂ reacts with 28 grams of N₂, 187.3 grams of H₂ reacts with how much mass of N₂?
[tex]mass of N_{2} =\frac{187.3 grams of H_{2} x28 grams of N_{2}}{6 grams of H_{2}}[/tex]
mass of N₂= 874.067 grams
But 875.087 grams of N₂ are not available, 50.8 grams are available. Since you have less mass than you need to react with 187.3 grams of H₂, N₂ will be the limiting reagent.
Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 28 grams of N₂ form 2 moles of NH₃, 50.8 grams of N₂ form how many moles of NH₃?
[tex]moles of NH_{3} =\frac{50.8 grams of N_{2} x2 moles of NH_{3} }{28 grams of N_{2}}[/tex]
moles of NH₃= 3.63 moles
Finally, 3.63 moles of NH₃ will be produced if 50.8 g N₂ reacts with 187.3 g H₂.
Learn more about the reaction stoichiometry:
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