how much aluminum oxide is needed to produce 1 kg of aluminum?
The reaction is
4Al + 3O2 -> 2 Al2O3
Two moles of Al2O3 are made by oxidizing four moles of Al.
1 kg of Al is 1000/27 = 37.0 moles. You will therefore make 18.5 moles of Al2O3. Convert that to kg if you wish.
To calculate the amount of aluminum oxide needed to produce 1 kg of aluminum, we need to use the molar ratio from the balanced chemical equation for the reaction of aluminum oxide with aluminum:
2 Al₂O₃ + 3 C ⟶ 4 Al + 3 CO₂
From this equation, we can see that for every 4 moles of aluminum produced, we need 2 moles of aluminum oxide.
The molar mass of aluminum oxide is calculated by adding the atomic masses of aluminum (Al = 26.98 g/mol) and oxygen (O = 16.00 g/mol) together.
Aluminum oxide (Al₂O₃) has a molar mass of:
2(Al) + 3(O) = 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol
Now, let's calculate the amount of aluminum oxide needed:
Moles of Aluminum oxide needed = (1 kg / molar mass of Al₂O₃)
= (1000 g / 101.96 g/mol)
≈ 9.81 mol
Since the molar ratio between aluminum and aluminum oxide is 4:2, we can divide the moles of aluminum oxide needed by 2 and multiply by 4 to find the moles of aluminum required.
Moles of Aluminum needed = (moles of Aluminum oxide needed / 2) x 4
= (9.81 mol / 2) x 4
= 19.62 mol
Therefore, approximately 9.81 moles of aluminum oxide are needed to produce 1 kg of aluminum.
To determine how much aluminum oxide is needed to produce 1 kg of aluminum, we need to understand the chemical reaction involved in the production process.
The production of aluminum typically involves the extraction of aluminum from its ore, which is usually bauxite. The main compound in bauxite is aluminum oxide (Al2O3). The extraction process is known as the Bayer process and involves several steps.
First, bauxite is crushed and mixed with a solution of sodium hydroxide (NaOH), which reacts with the aluminum oxide to form sodium aluminate (NaAlO2). This reaction can be represented as:
2 Al2O3 + 2 NaOH → 2 NaAlO2 + H2O
Next, the sodium aluminate solution is treated to separate the aluminum hydroxide (Al(OH)3) from impurities. Finally, the aluminum hydroxide undergoes further processing to produce pure aluminum.
Now, let's calculate the molar mass of aluminum oxide (Al2O3) using the atomic masses from the periodic table:
Molar mass of Al = 26.98 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Al2O3 = (2 × Molar mass of Al) + (3 × Molar mass of O)
= (2 × 26.98 g/mol) + (3 × 16.00 g/mol)
= 101.96 g/mol
Thus, the molar mass of aluminum oxide is approximately 101.96 g/mol.
To find the amount of aluminum oxide needed to produce 1 kg (1000 g) of aluminum, we use stoichiometry. According to the balanced chemical equation, 2 moles of aluminum oxide (Al2O3) react to produce 4 moles of aluminum (Al). Therefore, the ratio of moles of aluminum oxide to moles of aluminum is 2:4, or 1:2.
Now, we can calculate the moles of aluminum oxide needed:
Moles of aluminum oxide = Moles of aluminum × (1 mol of aluminum oxide / 2 mol of aluminum)
= 1 kg of aluminum × (1000 g / 26.98 g/mol) × (1 mol of aluminum oxide / 2 mol of aluminum)
≈ 18.53 mol
Finally, we can calculate the mass of aluminum oxide needed:
Mass of aluminum oxide = Moles of aluminum oxide × Molar mass of aluminum oxide
= 18.53 mol × 101.96 g/mol
≈ 1,891 g
Therefore, approximately 1,891 grams (or 1.891 kg) of aluminum oxide is needed to produce 1 kg of aluminum.