10.0 g aluminum reacts with 66.5g bromine to form 65.0g aluminum bromide. determine the limiting reactant and calculate the theoretical and percent yield. once the reaction has occured as completely as possible, what mass of the excess reactant is left?
To determine the limiting reactant and calculate the theoretical and percent yield, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction.
2 Al + 3 Br2 → 2 AlBr3
Step 2: Calculate the molar masses of aluminum (Al) and bromine (Br2).
Aluminum (Al) molar mass = 26.98 g/mol
Bromine (Br2) molar mass = 2 × 79.90 g/mol = 159.80 g/mol
Step 3: Calculate the number of moles for each reactant.
Number of moles of Al = mass of Al / molar mass of Al
Number of moles of Al = 10.0 g / 26.98 g/mol = 0.37 mol
Number of moles of Br2 = mass of Br2 / molar mass of Br2
Number of moles of Br2 = 66.5 g / 159.80 g/mol = 0.42 mol
Step 4: Determine the limiting reactant.
To determine the limiting reactant, we compare the mole ratios of Al and Br2 in the balanced equation. The coefficient ratio is 2:3.
Since the mole ratio of Al:Br2 is 2:3, the number of moles of Al required to react with the available moles of Br2 would be:
(0.42 mol of Br2) × (2 mol Al / 3 mol Br2) = 0.28 mol Al
Since the actual moles of Al available (0.37 mol) is greater than the required moles of Al (0.28 mol), Al is in excess, while Br2 is the limiting reactant.
Step 5: Calculate the theoretical yield.
The theoretical yield is the maximum amount of product that can be formed based on the limiting reactant. In this case, we'll calculate the theoretical yield of aluminum bromide (AlBr3).
The molar mass of AlBr3 = (1 × Al molar mass) + (3 × Br molar mass)
Molar mass of AlBr3 = (1 × 26.98 g/mol) + (3 × 79.90 g/mol)
Molar mass of AlBr3 = 266.76 g/mol
Theoretical yield of AlBr3 = moles of limiting reactant (Br2) × molar mass of AlBr3
Theoretical yield of AlBr3 = 0.42 mol × 266.76 g/mol = 112.03 g
Step 6: Calculate the percent yield.
Percent yield = (actual yield / theoretical yield) × 100
Given: actual yield = 65.0 g
Percent yield = (65.0 g / 112.03 g) × 100 = 58.03%
Step 7: Calculate the mass of the excess reactant remaining.
The excess reactant is aluminum (Al).
Mass of excess Al = (moles of Al - moles of Al used) × molar mass of Al
Mass of excess Al = (0.37 mol - 0.28 mol) × 26.98 g/mol = 2.42 g
Therefore, after the reaction has occurred, there will be 2.42 grams of excess aluminum (Al) remaining.