If 5.0 of al react with 9.0 g of O2, how many grams of Al2O3 (101.56 g/mol) can be formed

9.394 g of Al₂O₃

Explanation:

The reaction between Aluminium and Oxygen is given by the balanced equation;

4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)

We are given;

Mass of Aluminium as 5.0 g Mass of Oxygen as 9.0 g

We are required to determine the mass of Al₂O₃ formed.

We are going to use the following simple steps;

Step 1: Calculate the number of moles of Aluminium and Oxygen

Moles = Mass : Molar mass

Moles of Al;

Molar mass of Al = 26.98 g/mol

Moles of Al = 5.0 g : 26.98 g/mol

= 0.185 moles

Moles of Oxygen gas

Molar mass of O₂ = 32.0 g/mol

Moles O₂ = 9.0 g : 32.0 g/mol

= 0.28125 moles

From the equation, four moles of Aluminium reacts with three moles of oxygen gas. Therefore, Aluminium is the rate limiting reagent while oxygen gas is in excess. Step 2: Calculate the number of moles of Al₂O₃ formed

4 moles of Al reacts to produce 2 moles of Al₂O₃

Thus, the mole ratio of Al to Al₂O₃ is 2 : 1

Thus, moles of Al₂O₃ = Moles of Al : 2

= 0.185 moles : 2

= 0.0925 moles

Step 3: Calculate the mass of Al₂O₃ formed

Mass = Moles * Molar mass

Molar mass of Al₂O₃ is 101.56 g/mol

= 0.0925 moles * 101.56 g/mol

= 9.394 g

Therefore, the mass of Al₂O₃ formed is 9.394 g