<h3>Answer:</h3>
9.394 g of Al₂O₃
<h3>
Explanation:</h3>
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;
<h3>Step 1: Calculate the number of moles of Aluminium and Oxygen </h3>
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.
<h3>Step 2: Calculate the number of moles of Al₂O₃ formed </h3>
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
<h3>Step 3: Calculate the mass of Al₂O₃ formed</h3>
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
