JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Some Basic Concepts of Chemistry

The reaction is $4\mathrm{Al}+3\mathrm{O}{}_2\to 2\mathrm{Al}{}_2\mathrm{O}{}_3$. Moles of Al $=81.0/27.0=3$; moles of $\mathrm{O}{}_2$ $=128.0/32.0=4$. The stoichiometric ratio needs 3 mol $\mathrm{O}{}_2$ per 4 mol Al, so 3 mol Al requires only 2.25 mol $\mathrm{O}{}_2$; since 4 mol $\mathrm{O}{}_2$ is available, aluminium is the limiting reagent. From the equation, 4 mol Al gives 2 mol $\mathrm{Al}{}_2\mathrm{O}{}_3$, so 3 mol Al gives $3\times \frac{2}{4}=1.5$ mol $\mathrm{Al}{}_2\mathrm{O}{}_3$. With molar mass $2(27)+3(16)=102\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the mass formed is $1.5\times 102=153\mathrm{g}$.