JEE Main 2025 — Some Basic Concepts Of Chemistry Question with Solution

To determine the mass of aluminum oxide produced, we first consider the chemical reaction between aluminum and oxygen:

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

Given:

Molar mass of Al = 27.0 g/mol

Molar mass of O = 16.0 g/mol

81.0 g of Al and 128.0 g of O${}_2$

Calculating moles of reactants:

For aluminum (Al), using its molar mass:

$\frac{81\text{ g}}{27\text{ g/mol}}=3\text{ moles of Al}$

For oxygen (O${}_2$), using its molar mass (O${}_2$ is composed of two oxygen atoms):

$\frac{128\text{ g}}{32\text{ g/mol}}=4{\text{ moles of O}}_2$

Identifying the limiting reagent:

From the stoichiometry of the reaction:

4 moles of Al react with 3 moles of O${}_2$

3 moles of Al require $\frac{3}{4}\times 3$ moles of O${}_2$ = 2.25 moles of O${}_2$

Since only 2.25 moles of O${}_2$ are needed by 3 moles of Al, and we have 4 moles of O${}_2$, aluminum is the limiting reagent.

Calculating moles of ${\text{Al}}_2{\text{O}}_3$ formed:

From the reaction, 4 moles of Al yield 2 moles of ${\text{Al}}_2{\text{O}}_3$. Thus:

$\frac{3}{4}\times 2=\frac{3}{2}\text{ moles of }{\text{Al}}_2{\text{O}}_3$

Calculating mass of ${\text{Al}}_2{\text{O}}_3$:

The molar mass of ${\text{Al}}_2{\text{O}}_3$ is:

$(2\times 27)+(3\times 16)=102\text{ g/mol}$

So, the mass of $\frac{3}{2}$ moles of ${\text{Al}}_2{\text{O}}_3$ is:

$\frac{3}{2}\times 102=153\text{ grams}$

Thus, the mass of aluminum oxide produced is 153 grams.