Taking into account the reaction stoichiometry and limiting reagent, 9 grams of NO can be produced from 12 g of NH₃ and 12 g of O₂.
In first place, the balanced reaction is:
4 NH₃ + 5 O₂ → 4 NO + 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:
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 68 grams of NH₃ reacts with 160 grams of O₂, 12 grams of NH₃ reacts with how much mass of O₂?
[tex]mass of O_{2} =\frac{12 grams of NH_{3}X160 grams of O_{2} }{68 grams of NH_{3}}[/tex]
mass of O₂= 28.23 grams
But 28.23 grams of O₂ are not available, 12 grams are available. Since you have less mass than you need to react with 12 grams of NH₃, O₂ will be the limiting reagent.
Considering the limiting reagent, he following rule of three can be applied: if by reaction stoichiometry 160 grams of O₂ form 120 grams of NO, 12 of O₂ form how much mass of NO?
[tex]mass of NO=\frac{12 grams of O_{2} x120 grams of NO}{160 grams of O_{2}}[/tex]
mass of NO= 9 grams
Then, 9 grams of NO can be produced from 12 g of NH₃ and 12 g of O₂.
Learn more about the reaction stoichiometry:
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