Question

The volume of $C{O}_2$ liberated at STP when 10 g of 90% pure lime is heated completely is:

• A. $2.016l$
• B. $20.16l$
• C. $2.24l$
• D. $22.4l$

Answer 1

The correct option is A $2.016l$

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