To find the mass of aluminium oxide (Al2O3) produced, we start with the balanced chemical equation for the reaction:
4Al + 3O2 → 2Al2O3
1. **Molar Mass Calculation:**
- Al (Aluminium) = 27 g/mol
- O2 (Oxygen) = 32 g/mol
- Al2O3 (Aluminium oxide) = (2×27 + 3×16) g/mol = 102 g/mol
2. **Determine Limiting Reagent:**
The moles of Al from 81.0 g:
n(Al) = 81.0 g / 27 g/mol = 3.0 mol
The moles of O2 from 128.0 g:
n(O2) = 128.0 g / 32 g/mol = 4.0 mol
According to the equation, 4 moles of Al react with 3 moles of O2. Calculate the required oxygen for 3.0 mol of Al:
Required O2 = (3.0 mol Al) × (3/4) = 2.25 mol O2
Since 2.25 mol of O2 is less than the available 4.0 mol, Al is the limiting reagent.
3. **Calculate Mass of Al2O3 Produced:**
From 3.0 mol of Al (1.5 mol of Al2O3):
n(Al2O3) = 3.0 mol × (2/4) = 1.5 mol
Mass of Al2O3 = 1.5 mol × 102 g/mol = 153.0 g
The mass of aluminium oxide produced is 153 g.
This value falls within the expected range of 153 to 153 g.
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
Which of the following best represents the temperature versus heat supplied graph for water, in the range of \(-20^\circ\text{C}\) to \(120^\circ\text{C}\)? 