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