Question

Aluminum reacts with oxygen to produce aluminum oxide.

4Al(s) + 3O₂(g) → 2Al₂O₃(s)
A mixture of 82.49 g of aluminum (м = 26.98 g/mol) and 117.65 g of oxygen (м = 32.00 g/mol) react according to this equation. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

a) Aluminum is limiting; 15.38 g of oxygen excess
b) Oxygen is limiting; 22.75 g of aluminum excess
c) Aluminum is limiting; 22.75 g of oxygen excess
d) Oxygen is limiting; 15.38 g of aluminum excess

Answer 1

The limiting reactant is aluminum and the mass of the excess oxygen is 22.75 g. Option c.

Step-by-step explanation:

To determine the limiting reactant in a chemical reaction, we need to compare the amount of product that can be formed from each reactant. In this case, we are given the masses of aluminum and oxygen. We first need to calculate the number of moles of each reactant by dividing the mass of each by their respective molar masses.

For aluminum, we have 82.49 g / 26.98 g/mol = 3.057 mol. For oxygen, we have 117.65 g / 32.00 g/mol = 3.677 mol.

Next, we look at the stoichiometric ratio of aluminum to oxygen in the balanced equation, which is 4:3. This means that for every 4 moles of aluminum, we need 3 moles of oxygen to react completely. Since we have more moles of oxygen than aluminum, oxygen is in excess.

The limiting reactant is the reactant that determines how much product can be formed. In this case, the aluminum is the limiting reactant because we have less of it compared to the stoichiometric ratio.

Finally, we can calculate the mass of the excess reactant by subtracting the mass of the limiting reactant consumed from the total mass of the excess reactant. The molar mass of aluminum is 26.98 g/mol, so the mass of aluminum consumed is 3.057 mol * 26.98 g/mol = 82.529 g.

Therefore, the mass of the excess oxygen is 117.65 g - 82.529 g = 22.75 g.

So Option c is correct.