The balanced chemical equation for the reaction between aluminum and oxygen to produce aluminum oxide is:
4Al + 3O2 → 2Al2O3
This equation tells us that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.
To solve the problem, we need to determine which reactant is limiting and which is in excess. We can do this by calculating the number of moles of each reactant and comparing them to the stoichiometric ratio in the balanced equation.
The molar mass of aluminum is 26.98 g/mol, so 30 g of aluminum is equal to:
30 g Al × (1 mol Al / 26.98 g Al) = 1.11 mol Al
The molar mass of oxygen is 32.00 g/mol, so 26 g of oxygen is equal to:
26 g O2 × (1 mol O2 / 32.00 g O2) = 0.81 mol O2
According to the balanced equation, 4 moles of aluminum react with 3 moles of oxygen. Therefore, if all of the oxygen reacts, the maximum amount of aluminum that can react is:
3 mol O2 × (4 mol Al / 3 mol O2) = 4 mol Al
Since we have only 1.11 mol of aluminum, it is the limiting reactant. This means that all of the aluminum will react, and some of the oxygen will be left over.
The amount of aluminum oxide produced can be calculated using the stoichiometry of the balanced equation. Since 4 moles of aluminum produce 2 moles of aluminum oxide, then 1.11 moles of aluminum will produce:
(2 mol Al2O3 / 4 mol Al) × 1.11 mol Al = 0.555 mol Al2O3
The molar mass of aluminum oxide is 101.96 g/mol, so the mass of aluminum oxide produced is:
0.555 mol Al2O3 × (101.96 g Al2O3 / 1 mol Al2O3) = 56.5 g Al2O3
Therefore, 56.5 grams of aluminum oxide will be produced.