An excess of Al and 7.9 Mole of Br2 are reacted according to the question 2Al+3 Br2 equals 2AlBr3
To determine the limiting reactant and the theoretical yield of the product, you need to compare the number of moles of each reactant to their stoichiometric ratios.
The balanced chemical equation for the reaction is:
2Al + 3Br2 → 2AlBr3
From the equation, you can see that 2 moles of Al react with 3 moles of Br2 to produce 2 moles of AlBr3.
Given that there is an excess of Al and 7.9 moles of Br2, you can calculate the moles of AlBr3 that can be formed from each reactant.
1. The moles of AlBr3 that can be formed from Al:
Since Al is in excess, the moles of AlBr3 formed will depend on the amount of Br2.
Based on the stoichiometry, 3 moles of Br2 react with 2 moles of AlBr3.
Therefore, you can calculate the moles of AlBr3 formed from Br2:
Moles of AlBr3 from Br2 = (7.9 moles Br2) × (2 moles AlBr3 / 3 moles Br2)
2. The moles of AlBr3 that can be formed from Br2:
As Br2 is the limiting reactant, the moles of AlBr3 formed will be based on the amount of Br2.
From the stoichiometry, 3 moles of Br2 react with 2 moles of AlBr3.
Therefore, moles of AlBr3 formed from Br2 = 7.9 moles Br2 × (2 moles AlBr3 / 3 moles Br2)
Now you can compare the moles of AlBr3 formed from each reactant to find the limiting reactant and the theoretical yield.
The limiting reactant is the one that produces the smallest amount of AlBr3.
After comparing the moles of AlBr3 formed from Al and Br2, you will find that the moles of AlBr3 formed from Br2 are smaller than those formed from Al. Hence, Br2 is the limiting reactant.
To calculate the theoretical yield of AlBr3, you can use the moles of AlBr3 formed from Br2:
Theoretical yield of AlBr3 = moles of AlBr3 from Br2 × molar mass of AlBr3
Remember to look up the molar mass of AlBr3 and use proper units in your calculation.
Please note that this calculation assumes the reaction goes to completion and that there are no side reactions or losses during the process.
To determine the limiting reactant and the amount of excess reactant remaining in a chemical reaction, we can follow these steps:
Step 1: Convert the given amounts of reactants into moles.
Given:
- Excess Al
- 7.9 moles of Br2
Since we don't have a specific amount for the excess Al, we only need to convert the given amount for Br2 into moles. We can use the molar mass of Br2, which is 159.808 g/mol.
Moles of Br2 = Mass of Br2 / Molar mass of Br2
Moles of Br2 = 7.9 moles (given)
Step 2: Determine the stoichiometric ratio between the reactants.
From the balanced chemical equation: 2Al + 3Br2 -> 2AlBr3
We can see that the ratio of Al to Br2 required for complete reaction is 2:3.
Step 3: Calculate the moles of Al required for the given moles of Br2.
According to the stoichiometry, for every 3 moles of Br2, we need 2 moles of Al.
Moles of Al = (Moles of Br2) x (2 moles Al / 3 moles Br2)
Moles of Al = 7.9 moles (Br2) x (2 moles Al / 3 moles Br2)
Moles of Al = 5.27 moles
Step 4: Compare the actual moles of Al with the moles required based on the stoichiometry.
Since we have an excess of Al, the moles of Al we have (excess) is greater than the moles required. Therefore, Al is the excess reactant, and Br2 is the limiting reactant.
Step 5: Determine the amount of excess reactant remaining.
To find the amount of excess Al remaining, we subtract the moles of Al required from the moles of Al provided.
Excess Al remaining = Moles of Al (excess) - Moles of Al (required)
Excess Al remaining = Excess Al - 5.27 moles
Note: If you have the amount of excess Al in grams, you will need to convert it to moles by dividing by the molar mass of Al (26.981539 g/mol).
I hope this explanation helps! Let me know if you have any further questions.