Respuesta :
Taking into account the reaction stoichiometry and the definition of limiting reaction, 528.5 grams of AlI₃ are formed when 35 g of Al reacts with 495 g of I₂.
Reaction stoichiometry
In first place, the balanced reaction is:
2 Al + 3 I₂ → 2 AlO₃
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:
- Al: 2 moles
- I₂: 3 moles
- AlI₃: 2 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- I₂: 253.8 g/mole
- AlI₃: 407.7 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 2 moles× 27 g/mole= 54 grams
- I₂: 3 moles× 253.8 g/mole= 761.4 grams
- AlI₃: 2 moles× 407.7 g/mole= 815.4 grams
Limiting reagent
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 the reaction stoichiometry and a simple rule of three as follows: if by stoichiometry 761.4 grams of I₂ reacts with 54 grams of Al, if 495 grams of I₂ react, how much mass of Al will be needed?
[tex]mass of Al=\frac{495 grams of I_{2}x 54 grams of Al }{761.4 gramd of I_{2} } [/tex]
mass of Al=35.1 grams
But 35.1 grams of Al are not available, 35 grams are available. Since you have less mass than you need to react with 495 grams of I₂, Al will be the limiting reagent.
Mass of AlI₃ formed
The following rule of three can be applied: if by reaction stoichiometry 54 grams of Al form 815.4 grams of AlI₃, 35 grams of Al form how much mass of AlI₃?
[tex]mass of AlI_{3} =\frac{35 grams of Alx815.4 grams of AlI_{3}}{54 grams of Al} [/tex]
mass of AlI₃= 528.5 grams
Finally, 528.5 grams of AlI₃ are formed when 35 g of Al reacts with 495 g of I₂.
Learn more about reaction stoichiometry:
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