Aluminum reacts spontaneously with bromine to produce AlBr3.
If 218.08 grams of aluminum reacts with 258.21 grams of bromine and the reaction produces 103.74 grams of AlBr3, what is the percent yield for the reaction?
This problem is solved the same way as the Sb2S3 problem. Just follow that process. Post your work if you get stuck.
To find the percent yield, we first need to calculate the theoretical yield, which represents the maximum amount of AlBr3 that could be produced according to the balanced chemical equation.
The balanced chemical equation for the reaction between aluminum and bromine is:
2 Al + 3 Br2 -> 2 AlBr3
From the equation, we can see that 2 moles of aluminum react with 3 moles of bromine to produce 2 moles of AlBr3.
Step 1: Convert the given masses of aluminum and bromine to moles.
Molar mass of aluminum (Al) = 26.98 g/mol
Molar mass of bromine (Br2) = 79.90 g/mol
Number of moles of aluminum = mass of aluminum / molar mass of aluminum
Number of moles of aluminum = 218.08 g / 26.98 g/mol = 8.09 mol (rounded to two decimal places)
Number of moles of bromine = mass of bromine / molar mass of bromine
Number of moles of bromine = 258.21 g / 79.90 g/mol = 3.23 mol (rounded to two decimal places)
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the mole ratio of aluminum to bromine in the balanced equation to the actual mole ratio calculated from the given masses.
The mole ratio of aluminum to bromine in the balanced equation is 2:3. This means that for every 2 moles of aluminum, we need 3 moles of bromine.
The actual mole ratio calculated from the given masses is:
Aluminum: Number of moles of aluminum = 8.09 mol
Bromine: Number of moles of bromine = 3.23 mol
Dividing the number of moles of bromine by the balanced mole ratio gives us the expected number moles of aluminum needed:
Number of moles of bromine / mole ratio = 3.23 mol / 3 = 1.08 mol
Since the actual number of moles of aluminum (8.09 mol) is greater than the expected number of moles needed (1.08 mol), aluminum is the excess reactant, and bromine is the limiting reactant.
Step 3: Calculate the theoretical yield.
To determine the theoretical yield, we need to calculate the number of moles of AlBr3 that can be produced based on the limiting reactant (bromine).
From the balanced equation, we know that:
2 moles of AlBr3 are produced from 3 moles of Br2
Using the mole ratio, we can calculate the number of moles of AlBr3 produced:
Number of moles of AlBr3 = (3.23 mol Br2) × (2 mol AlBr3 / 3 mol Br2) = 2.15 mol (rounded to two decimal places)
Step 4: Calculate the percent yield.
The percent yield is calculated by dividing the actual yield (103.74 g) by the theoretical yield (calculated in Step 3) and multiplying by 100.
Percent yield = (Actual yield / Theoretical yield) × 100
Percent yield = (103.74 g / 2.15 mol) × 100 = 48.24% (rounded to two decimal places)
Therefore, the percent yield for the reaction is approximately 48.24%.