When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is ----- (nearest integer).
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Correct Answer: 153
Approach Solution - 1
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.
Approach Solution -2
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