When 81.0\, g of Al reacts with 128.0\, g of O 2 , find the mass of Al 2O 3 formed (nearest integer). (Given: molar mass of Al =27.0\, g\,mol -1 and molar mass of O =16.0\, g\,mol -1 .)

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 23 January, Evening Shift — Question 45

When $81.0\,\mathrm{g}$ of $\mathrm{Al}$ reacts with $128.0\,\mathrm{g}$ of $\mathrm{O_2}$ , find the mass of $\mathrm{Al_2O_3}$ formed (nearest integer).

(Given: molar mass of $\mathrm{Al}=27.0\,\mathrm{g\,mol^{-1}}$ and molar mass of $\mathrm{O}=16.0\,\mathrm{g\,mol^{-1}}$ .)

Answer: 153

Numerical answer — enter this value.

Step-by-step solution

4\,\mathrm{Al} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{Al_2O_3}

n(\mathrm{Al}) = \frac{81.0}{27.0} = 3.0\,\mathrm{mol}

n(\mathrm{O_2}) = \frac{128.0}{32.0} = 4.0\,\mathrm{mol}

4\,\mathrm{Al} \rightarrow 2\,\mathrm{Al_2O_3}

3\,\mathrm{Al} \rightarrow \frac{2}{4}\times 3 = 1.5\,\mathrm{mol\,Al_2O_3}

M(\mathrm{Al_2O_3}) = 2(27) + 3(16) = 102\,\mathrm{g\,mol^{-1}}

m(\mathrm{Al_2O_3}) = 1.5 \times 102 = 153\,\mathrm{g}

\text{Answer} = 153

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Chemical Equations, Stoichiometry and Limiting Reagent

0.01 mole of an organic compound (X) containing 10\% hydrogen, on complete combustion, produced 0.9 g of H 2O . Find the molar mass of (X)…

Answer: 153

Step-by-step solution

Practise Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

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