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) Given : Molar mass of Al is 27.0 g $$mol^{-1}$$ Molar mass of O is 16.0 g $$mol^{-1}$$.

81 g Al reacts with 128 g O₂. Find mass of Al₂O₃ produced.

Write the balanced equation

$$4Al + 3O_2 \rightarrow 2Al_2O_3$$

Find moles

Moles of Al = 81/27 = 3 mol

Moles of O₂ = 128/32 = 4 mol

Find limiting reagent

For 3 mol Al, we need $$\frac{3}{4} \times 3 = 2.25$$ mol O₂. We have 4 mol O₂.

So Al is the limiting reagent.

Calculate product

3 mol Al produces $$\frac{2}{4} \times 3 = 1.5$$ mol Al₂O₃

Mass = 1.5 × 102 = 153 g

The answer is 153.