1. Write the balanced chemical equation

$4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$

2. Calculate moles of reactants

For aluminium:

$n(\text{Al}) = \frac{81.0}{27.0} = 3.0\ \text{mol}$

For oxygen gas:

Molar mass of $\text{O}_2 = 2\times 16.0 = 32.0\ \text{g mol}^{-1}$

$n(\text{O}_2) = \frac{128.0}{32.0} = 4.0\ \text{mol}$

3. Find the limiting reagent

From the balanced equation:

$4\ \text{mol Al} \text{ react with } 3\ \text{mol O}_2$

So, for $3.0$ mol Al, required oxygen is:

$\text{Required } \text{O}_2 = 3.0 \times \frac{3}{4} = 2.25\ \text{mol}$

Available oxygen = $4.0$ mol, which is excess.

Therefore, Al is the limiting reagent.

4. Calculate moles of aluminium oxide formed

From the equation:

$4\ \text{mol Al} \rightarrow 2\ \text{mol Al}_2\text{O}_3$

Thus,

$3.0\ \text{mol Al} \rightarrow 3.0 \times \frac{2}{4} = 1.5\ \text{mol Al}_2\text{O}_3$

5. Calculate molar mass of aluminium oxide

$M(\text{Al}_2\text{O}_3) = 2(27.0) + 3(16.0) = 54 + 48 = 102\ \text{g mol}^{-1}$

6. Calculate mass of aluminium oxide produced

$m = n \times M = 1.5 \times 102 = 153\ \text{g}$

7. Nearest integer

$\boxed{153}$

The derived answer matches the stored correct answer.