Respuesta :
Answer:
30.17 grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al.
19.33 grams of the excess react will remain once the reaction goes to completion.
Explanation:
You know:
2 Al + Cr₂O₃ → Al₂O₃ + 2 Cr
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of the compounds participating in the reaction react and are produced:
- Al: 2 moles
- Cr₂O₃: 1 mole
- Al₂O₃: 1 mole
- Cr: 2 moles
Being:
- Al: 27 g/mole
- Cr: 52 g/mole
- O: 16 g/mole
the molar mass of the compounds participating in the reaction is:
- Al: 27 g/mole
- Cr₂O₃: 2*52 g/mole + 3 *16 g/mole= 152 g/mole
- Al₂O₃: 2*27 g/mole + 3 *16 g/mole= 102 g/mole
- Cr: 52 g/mole
Then, by stoichiometry of the reaction, the amounts of reagents and products that participate in the reaction are:
- Al: 2 moles* 27 g/mole= 54 g
- Cr₂O₃: 1 mole* 152 g/mole= 152 g
- Al₂O₃: 1 mole* 102 g/mole= 102 g
- Cr: 2moles*52 g/mole= 104 g
First, you must determine the 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.
Then, for this you must apply the following rule of three: if 54 grams of Al react with 152 grams of Cr₂O₃ by stoichiometry of the reaction, 35 grams of Al with how much mass of Cr₂O₃ will react?
[tex]mass of Cr_{2} O_{3} =\frac{35 grams of Al*152 gramsof Cr_{2} O_{3}}{54 grams of Al}[/tex]
mass of Cr₂O₃= 98.52 grams
But 98.52 grams of Cr₂O₃ are not available, 44.1 grams are available. Since you have less moles than you need to react with 35 grams of Al, the compound Cr₂O₃ will be the limiting reagent.
To determine the amount of Cr that can be produced, you use the amount of limiting reagent available and apply the following rule of three: if by stoichiometry 152 grams of Cr₂O₃ produce 104 grams of Cr, 44.1 grams of Cr₂O₃ how much mass of Cr does it produce?
[tex]mass of Cr =\frac{104 grams of Cr*44.1 grams of Cr_{2} O_{3}}{152 grams of Cr_{2} O_{3}}[/tex]
mass of Cr= 30.17 grams
30.17 grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al.
As Al is the excess reagent, you must first calculate the amount of mass that reacts with 44.1 grams of Cr₂O₃ using the following rule of three: if 152 grams of Cr₂O₃ react with 54 grams of Al, 44.1 grams of Cr₂O₃ with how much mass of Al reaction to?
[tex]mass of Al =\frac{54 grams of Al*44.1 gramsnof Cr_{2} O_{3}}{152 gramsnof Cr_{2} O_{3}l}[/tex]
mass of Al=15.67 grams
With 35 grams being the amount of Al available, the amount of Al that will remain in the reaction after all the limiting reagent reacts and the reaction is complete is calculated by:
mass of excess= 35 grams - 15.67 grams= 19.33 grams
19.33 grams of the excess react will remain once the reaction goes to completion.
Stoichiochemistry is used to determine the number of moles reacting in a reaction to produce the corresponding number of moles.
The correct answers are:
30.17 grams of Chromium (Cr) can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Aluminum (Al).
19.33 grams of the excess reactant will be remained once the reaction goes to completion.
The reaction between aluminum and chromium oxide is:
2 Al + Cr₂O₃ → Al₂O₃ + 2 Cr
From the stoichiometry, it can be concluded that:
Al - 2 moles x 27 g/mole= 54 g
Cr₂O₃ - 1 mole x 152 g/mole= 152 g
Al₂O₃ - 1 mole x 102 g/mole= 102 g
Cr - 2moles x 52 g/mole= 104 g
The stoichiometry shows that for every 54 grams of Al, 152 grams of chromium oxide react. To calculate the amount of chromium oxide that will react with 35 grams of Al, we have:
[tex]\text{Mass of grams of Cr}_2 O_3 &=}\dfrac{35 \text{grams of Al} \times 152 \text{grams of Cr}_2 O_3}{54 \;\text{grams of Al}}[/tex]
Mass of Chromium oxide = 98.52 g
From the stoichiometry, 152 g of chromium oxide produce 104 g of Chromium. Now, to calculate the 44.1 g f chromium oxide produced, we have:
[tex]\text{Mass of Cr &=}\dfrac{104\;\text{grams of Al} \times 44.1 \text{grams of Cr}_2 O_3}{152 \;\text{grams of Cr}_2O_3}[/tex]
Mass of Cr = 30.17 grams
Now, we know that:
If 152 grams of chromium oxide react with 54 grams of Aluminum, then number of Aluminum required to react with 44.1 chromium oxide is:
[tex]\text{Mass of Al&=}\dfrac{54\;\text{grams of Al} \times 44.1 \text{grams of Cr}_2 O_3}{152 \;\text{grams of Cr}_2O_3}[/tex]
Mass of Al = 15.67 g
Thus, the amount of aluminum present in the reaction after the limiting reagent is utilized:
Excess mass = 35 - 15.67 = 19.33 g
Therefore, 19.33 g of Aluminum will be excess.
To know more about limiting reagent, refer to the following link:
https://brainly.com/question/20070272
