if an excess of Al and 7.9 mol of Br2 are reacted in equation 2Al + 3Br2= 2AlBr2, how many moles of AlBr2 will be formed assuming a 100% yield
I can't help but wonder if that AlBr2 is a typo meant to be AlBr3.
Well, buckle up, because we're about to get chemical!
From the balanced equation, we can see that 2 moles of Al react with 3 moles of Br2 to form 2 moles of AlBr2. So, the moles of AlBr2 formed will be directly proportional to the moles of Al used.
You stated that there is an excess of Al, so we don't have to worry about the Al running out before the Br2. Therefore, since we have 7.9 moles of Br2, we can assume that all of it will react completely with the Al.
Using the ratio from the equation, we can calculate the moles of AlBr2 formed:
(7.9 mol Br2) x (2 mol AlBr2 / 3 mol Br2) = 5.27 mol AlBr2
So, assuming a 100% yield (which rarely happens in real life, but hey, we're being optimistic here), approximately 5.27 moles of AlBr2 will be formed.
I hope that brings a smile to your face like a clown with a squeaky red nose!
To find out how many moles of AlBr2 will be formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed.
Let's start by calculating the moles of Al and Br2 available:
1) Moles of Al:
Given: Excess Al
Therefore, the moles of Al are not relevant for determining the limiting reactant.
2) Moles of Br2:
Given: 7.9 mol
From the balanced equation: 2Al + 3Br2 = 2AlBr2
The stoichiometric ratio of Br2 to AlBr2 is 3:2.
So, for each mole of Br2, we expect to form 2/3 moles of AlBr2.
To calculate the moles of AlBr2 formed, we will use the moles of Br2 available:
Moles of AlBr2 formed = (7.9 mol Br2) x (2 moles AlBr2 / 3 moles Br2)
= 5.27 mol
Therefore, assuming a 100% yield, 5.27 moles of AlBr2 will be formed.
To determine the number of moles of AlBr2 formed, we need to use the stoichiometry of the balanced equation.
The balanced equation is 2Al + 3Br2 → 2AlBr2, which shows that 2 moles of Al reacts with 3 moles of Br2 to produce 2 moles of AlBr2.
Since an excess of Al is given, it means that Al is in excess, and all of the Br2 will be completely consumed in the reaction.
Given that there are 7.9 mol of Br2 available, we can calculate the moles of AlBr2 formed using the stoichiometry of the balanced equation:
7.9 mol Br2 × (2 mol AlBr2 / 3 mol Br2) = 5.2667 mol AlBr2
Therefore, assuming a 100% yield, 5.2667 moles of AlBr2 will be formed in the reaction.
Tiffany, Assuming the reaction runs to completion and there is an excess of Al you know that Bromine is your limiting reagent.
Stoichiometry will give you the amount of AlBr2 produced.
You begin with 7.9 mol Br2 to cancel the moles of Bromine and find moles of Aluminum Bromide you must multiply by the mole ratio found in the balanced chemical equation:
7.9mol Br2 x (2molAlBr2/3molBr2) = ___ mol AlBr2
This yields 5.3 moles Aluminum Bromide.