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.