Question

The volume in litres of $C{O}_2$ liberated at STP when 10 grams of 90% pure limestone is heated completely is

• A. 22.4
• B. 2.24
• C. 20.6
• D. 2.016

Answer 1

Answer: (D)

$<br/>CaC{O}_3\stackrel{\Delta }{⟶}CaO+C{O}_2\uparrow <br/>$
The molar mass of $CaC{O}_3=100g/mol$.
$10g$ of $90\%$ pure lime $=\frac{10g}{100g/mol}\times \frac{90}{100}=0.09$ moles $CaC{O}_3$
$0.09$ moles $CaC{O}_3$ on heating gives $0.09$ moles $C{O}_2$.
At STP. 1 mole $C{O}_2=22.4L$.
At STP. $0.09$ mole $C{O}_2=0.09\times 22.4=2.016L$.
Hence, the volume of $C{O}_2$ liberated at STP when $10g$ of $90\%$ pure lime is heated completely is $2.016L$.