An excess of Al and 7.9 mole of Br2 are reacted according to the equation 2Al+ 3Br2 equals 2AlBr3. How many moles of AlBr3 will be formed assuming 100% is yield?
mols Br2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols Br2 to mols AlBr3.
To find out the number of moles of AlBr3 formed, we need to calculate the limiting reagent first.
Given:
- Excess Al
- 7.9 moles of Br2
The balanced chemical equation is:
2 Al + 3 Br2 -> 2 AlBr3
From the equation, we can see that the ratio of Al to Br2 is 2:3.
Let's calculate the moles of Al available:
Since Al is in excess, the number of moles of Al will not be a limiting factor in the reaction. Therefore, we only need to consider the number of moles of Br2.
The moles of Br2 available is 7.9 moles.
Now let's calculate the number of moles of AlBr3 formed using the moles of Br2:
For every 3 moles of Br2, we get 2 moles of AlBr3.
Using the ratio, we can calculate:
7.9 moles of Br2 * (2 moles of AlBr3 / 3 moles of Br2) = 5.27 moles of AlBr3.
Therefore, assuming 100% yield, 5.27 moles of AlBr3 will be formed.
To determine the number of moles of AlBr3 formed, we can start by calculating the limiting reactant in this chemical reaction. The limiting reactant is the reactant that is completely consumed, thereby determining the maximum amount of product that can be formed.
In this case, we have an excess of Al and 7.9 moles of Br2. We need to compare the number of moles of each reactant to determine which one is the limiting reactant.
According to the balanced equation: 2Al + 3Br2 → 2AlBr3, we can see that the stoichiometric ratio is 2:3 between Al and Br2. This means that for every 2 moles of Al, we need 3 moles of Br2 to completely react.
To calculate the moles of Al required to react with 7.9 moles of Br2, we use the stoichiometric ratio:
7.9 moles of Br2 * (2 moles of Al / 3 moles of Br2) = 5.27 moles of Al
Now, we compare this result with the initial moles of Al. Since there is an excess of Al, we can conclude that Al is not the limiting reactant. Therefore, the limiting reactant is Br2.
Since we know the number of moles of the limiting reactant, we can use the stoichiometry to determine the number of moles of AlBr3 produced. Again, using the stoichiometric ratio:
7.9 moles of Br2 * (2 moles of AlBr3 / 3 moles of Br2) = 5.27 moles of AlBr3
Therefore, assuming 100% yield, 5.27 moles of AlBr3 will be formed.