Respuesta :
Answer:
155.4 g of Al₂O₃
Explanation:
The reaction is:
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
To determine the mass of aluminum oxide that is formed we need to know the limiting reagent. Let's calculate the moles of each by the molar mass
Mass / molar mass = Moles
82.49 g / 26.98 g/mol = 3.05 moles of Al
117.65 g / 32 g/mol = 3.67 moles of oxygen
Let's try the oxygen. Ratio is 3:4.
3 moles of O₂ need 4 moles of Al to react
Therefore 3.67 moles of O₂ will react with (3.67 . 4 )/3 = 4.90 moles
We only have 3.05 moles of Al, so the Al is the limiting reactant
Now, we work with stoichiometry
4 moles of Al can produce 2 moles of Al₂O₃
3.05 moles of Al will produce (3.05 .2) / 4 = 1.52 moles of Al₂O₃
We convert the moles to mass: 1.52 mol . 101.96 g / 1mol = 155.4 g
Answer:
155.9 grams of aluminium oxide can be formed
Explanation:
Step 1: Data given
Mass of aluminium = 82.49 grams
Molar mass of aluminium = 26.98 g/mol
Mass of oxygen = 117.65 grams
Molar mass of oxygen = 32.0 g/mol
Molar mass of aluminium oxide = 101.96 g/mol
Step 2: The balanced equation
4Al + 3O2 → 2Al2O3
Step 3: Calculate moles
Moles = mass / molar mass
Moles aluminium = 82.49 grams / 26.98 g/mol
Moles aluminium = 3.057 moles
Moles oxygen = 117.65 grams / 32.0 g/mol
Moles oxygen = 3.677 moles
Step 4: Calculate limiting reactant
For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3
Aluminium is the limiting reactant. It will completely be consumed (3.057 moles). Oxygen is in excess. There will react 3/4 * 3.057 = 2.293 moles
There will remain 3.677 - 2.293 = 1.384 moles
Step 5: Calculate moles aluminium oxide
For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3
For 3.057 moles Al we'll have 3.057/2 = 1.529 moles
Step 6: Calculate mass aluminium oxide
Mass aluminium oxide = 1.529 moles * 101.96 g/mol
Mass aluminium oxide = 155.9 grams
155.9 grams of aluminium oxide can be formed
