Beryllium Plus flourine
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Taking into account the reaction stoichiometry, the theorical yield for the reaction is 8.0467 grams of NH₃.
<h3>Reaction stoichiometry</h3>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:
- H₂: 3 moles
- N₂: 1 mole
- NH₃: 2 moles
The molar mass of the compounds is:
- H₂: 2 g/mole
- N₂: 28 g/mole
- NH₃: 17 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- H₂: 3 moles ×2 g/mole= 6 grams
- N₂: 1 mole ×28 g/mole= 28 grams
- NH₃: 2 moles ×17 g/mole= 34 grams
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.
<h3>Limiting reagent in this case</h3>To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 28 grams of N₂ reacts with 6 grams of H₂, 10.4 grams of N₂ reacts with how much mass of H₂?
<u><em>mass of H₂= 2.2286 grams</em></u>
But 2.2286 grams of H₂ are not available, 1.42 grams are available. Since you have less mass than you need to react with 10.4 grams of N₂, H₂ will be the limiting reagent.
<h3>Definition of theorical yield</h3>The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
<h3>Theoretical yield in this case</h3>Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 34 grams of NH₃, 1.42 grams of H₂ form how much mass of NH₃?
<u><em>mass of NH₃= 8.0467 grams</em></u>
Then, the theorical yield for the reaction is 8.0467 grams of NH₃.
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Answer:
The number of moles of H₂O is 0.135 mol.
The number of moles of LiClO₄ is 0.0449 mol.
Explanation:
Mole is an important standard unit used for the measurement of large quantities of atoms, molecules, or other particles. One mole is equal to 6.022×10²³ units.
The number of moles of a substance is calculated by:
To find the number of moles of H₂O:
Mass of H₂O in the sample = 2.43g
The molecular weight of H₂O = 18.02g
Number of moles = = 0.135 mol.
To find the number of moles of LiClO₄:
Mass of LiClO₄ given = 4.78g
The molecular weight of LiClO₄ = 106.39g
Number of moles = = 0.0449 mol.
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