For the reaction shown, compute the theoretical yield of the product in grams for each of the following initial amounts of reactancts
2Al(s) + 3Cl2(g) = 2AlCl3(s)
You gave no details but here is what you do.
1. Convert grams of what to have to moles. moles = grams/molar mass.
2. Using the coefficients in the balanced equation convert moles of what you have to moles of what you want (the product).
3. Now convert moles of the product to grams.
grams = mols x molar mass.
To calculate the theoretical yield of the product (AlCl3) in grams, we need to use the stoichiometry of the reaction and the given amounts of reactants.
The balanced equation tells us that the ratio of aluminum (Al) to aluminum chloride (AlCl3) is 2:2, and the ratio of chlorine (Cl2) to aluminum chloride (AlCl3) is 3:2.
Let's consider each case separately:
1. If we have 4 grams of aluminum (Al):
Since the molar mass of aluminum is 27 g/mol, we can calculate the number of moles of Al using the formula:
moles of Al = mass of Al / molar mass of Al
moles of Al = 4 g / 27 g/mol
moles of Al ≈ 0.148 mol
Using the stoichiometric ratios, we find that we need an equal number of moles of AlCl3. Therefore, the number of moles of AlCl3 = 0.148 mol.
Now, to find the mass of AlCl3, we can use the formula:
mass of AlCl3 = moles of AlCl3 × molar mass of AlCl3
mass of AlCl3 = 0.148 mol × (molar mass of AlCl3)
The molar mass of AlCl3 can be calculated by adding up the molar masses of one aluminum (27 g/mol) and three chlorines (35.5 g/mol each) as follows:
molar mass of AlCl3 = (1 × molar mass of Al) + (3 × molar mass of Cl)
molar mass of AlCl3 = (1 × 27 g/mol) + (3 × 35.5 g/mol)
molar mass of AlCl3 = 133.5 g/mol
Now we can substitute the values into the equation:
mass of AlCl3 = 0.148 mol × 133.5 g/mol
mass of AlCl3 ≈ 19.758 g
Therefore, the theoretical yield of AlCl3 is approximately 19.758 grams when starting with 4 grams of aluminum.
2. If we have 6 grams of chlorine (Cl2):
Since the molar mass of chlorine is 35.5 g/mol, we can calculate the number of moles of Cl2 using the formula:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 6 g / 35.5 g/mol
moles of Cl2 ≈ 0.169 mol
According to the stoichiometry of the reaction, the ratio of Cl2 to AlCl3 is 3:2. Therefore, the number of moles of AlCl3 = (2/3) × moles of Cl2
moles of AlCl3 = (2/3) × 0.169 mol
moles of AlCl3 ≈ 0.113 mol
Using the molar mass of AlCl3 (133.5 g/mol), we can find the mass of AlCl3:
mass of AlCl3 = moles of AlCl3 × molar mass of AlCl3
mass of AlCl3 = 0.113 mol × 133.5 g/mol
mass of AlCl3 ≈ 15.105 g
Therefore, the theoretical yield of AlCl3 is approximately 15.105 grams when starting with 6 grams of chlorine.
To calculate the theoretical yield of the product in grams, you need to follow these steps:
Step 1: Start by writing a balanced chemical equation for the reaction. The balanced equation for the given reaction is:
2Al(s) + 3Cl2(g) → 2AlCl3(s)
This indicates that 2 moles of aluminum (Al) react with 3 moles of chlorine gas (Cl2) to produce 2 moles of aluminum chloride (AlCl3).
Step 2: Determine the molar mass of each substance involved in the reaction:
The molar mass of Al = 26.98 g/mol (from the periodic table)
The molar mass of Cl2 = 70.90 g/mol (from the periodic table)
The molar mass of AlCl3 = (26.98 g/mol x 2) + (35.45 g/mol x 3) = 133.34 g/mol
Step 3: Calculate the number of moles of reactants:
You are given the initial amounts of reactants, so you need to convert them into moles.
Given:
- Initial amount of Al = 5.0 g
- Initial amount of Cl2 = 10.0 g
To calculate the number of moles, use the formula:
moles = mass / molar mass
Moles of Al = 5.0 g / 26.98 g/mol = 0.185 mol
Moles of Cl2 = 10.0 g / 70.90 g/mol = 0.141 mol
Step 4: Determine the limiting reactant:
To determine the limiting reactant, compare the mole ratio of the reactants from the balanced equation.
From the balanced equation:
2 moles of Al reacts with 3 moles of Cl2
Therefore, for every 2 moles of Al, we need 3/2 moles of Cl2.
Since we have 0.185 moles of Al and 0.141 moles of Cl2:
(0.185 moles Al) * (3/2 moles Cl2 per 2 moles Al) = 0.278 moles Cl2 needed
Since we have less than the required 0.278 moles of Cl2, Cl2 is our limiting reactant.
Step 5: Calculate the theoretical yield of the product:
Now that we know Cl2 is the limiting reactant, we can calculate the theoretical yield of AlCl3, which is the same as the moles of Cl2 used.
Theoretical yield of AlCl3 = moles of Cl2 used * molar mass of AlCl3
Theoretical yield of AlCl3 = 0.141 mol * 133.34 g/mol = 18.78 g
Therefore, the theoretical yield of the product AlCl3 is 18.78 grams.