What mass of aluminum is produced by the decomposition of 5.0 kg of aluminum chloride?
5 kg AlCl3 x fraction of Al in AlCl3 =
5 kg AlCl3 x (atomic mass Al/molar mass AlCl3) = ? kg Al
To determine the mass of aluminum produced by the decomposition of aluminum chloride, we need to understand the chemical reaction involved. The balanced chemical equation for the decomposition of aluminum chloride is as follows:
2AlCl₃ → 2Al + 3Cl₂
From the equation, we can see that 2 moles of aluminum chloride (AlCl₃) produce 2 moles of aluminum (Al) and 3 moles of chlorine gas (Cl₂).
First, we need to convert the mass of aluminum chloride given (5.0 kg) into moles. We do this by using the molar mass of aluminum chloride, which can be calculated by adding up the atomic masses of aluminum (Al) and chlorine (Cl):
Molar mass of AlCl₃ = (1 x atomic mass of Al) + (3 x atomic mass of Cl)
= (1 x 26.98 g/mol) + (3 x 35.45 g/mol)
= 133.33 g/mol
Next, we can calculate the number of moles of aluminum chloride:
Moles of AlCl₃ = Mass of AlCl₃ / Molar mass of AlCl₃
= 5000 g / 133.33 g/mol
≈ 37.50 mol
Since the stoichiometry of the reaction tells us that 2 moles of AlCl₃ produce 2 moles of Al, we can determine the number of moles of aluminum produced:
Moles of Al = Moles of AlCl₃
= 37.50 mol
Finally, we can calculate the mass of aluminum produced using the molar mass of aluminum (Al):
Mass of Al = Moles of Al x Molar mass of Al
= 37.50 mol x 26.98 g/mol
≈ 1,011.15 g
Therefore, the mass of aluminum produced by the decomposition of 5.0 kg of aluminum chloride is approximately 1,011.15 grams.