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Q.
The volume in litres of $CO_2$ liberated at STP when 10 grams of 90% pure limestone is heated completely is
Some Basic Concepts of Chemistry
Solution:
$
CaCO _{3} \stackrel{\Delta}{\longrightarrow} CaO + CO _{2} \uparrow
$
The molar mass of $CaCO _{3}=100 g / mol$.
$10 g$ of $90 \%$ pure lime $=\frac{10 g }{100 g / mol } \times \frac{90}{100}=0.09$ moles $CaCO _{3}$
$0.09$ moles $CaCO _{3}$ on heating gives $0.09$ moles $CO _{2}$.
At STP. 1 mole $CO _{2}=22.4 L$.
At STP. $0.09$ mole $CO _{2}=0.09 \times 22.4=2.016 L$.
Hence, the volume of $CO _{2}$ liberated at STP when $10 g$ of $90 \%$ pure lime is heated completely is $2.016 L$.