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

JEE Main 2025 — 4 April, Evening Shift — Question 16

The amount of calcium oxide produced on heating $150\,\mathrm{kg}$ limestone $(75\%$ pure $)$ is $\underline{\hspace{1cm}}$ kg. (Nearest integer)

Given: Molar mass (in g mol $^{-1}$ ): $\mathrm{Ca}=40,\ \mathrm{O}=16,\ \mathrm{C}=12$

Answer: 63

Numerical answer — enter this value.

Mass of pure $\mathrm{CaCO}_{3}=\frac{150 \times 75}{100}=112.5 \mathrm{~kg}$

Number of moles $=\frac{112.5}{100} \times 10^{3}=1125$ moles

Thermal decomposition,

$\mathrm{CaCO}_{3}(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$

Moles of CaO formed $=1125\; \mathrm{mol}$

Mass of $\mathrm{CaO}=\frac{1125 \times 56}{1000}=63\; \mathrm{kg}$

Answer key and solution verified before publishing.

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