Chemistry - Mole Concept Question with Solution | TestHub

Mass of pure ${\text{CaCO}}_3$ in the sample is $500\times \frac{90}{100}=450$ g.

Molar mass of ${\text{CaCO}}_3=40+12+3\times 16=100$ g/mol.

Moles of ${\text{CaCO}}_3=\frac{450}{100}=4.5$ mol.

The decomposition reaction is:

${\text{CaCO}}_3\to \text{CaO}+{\text{CO}}_2$

From the stoichiometry, $1$ mole of ${\text{CaCO}}_3$ produces $1$ mole of $\text{CaO}$ (lime).

Moles of $\text{CaO}$ produced $=4.5$ mol.

Molar mass of $\text{CaO}=40+16=56$ g/mol.

Mass of lime obtained $=4.5\times 56=252$ g.

Answer: $252$ g