Question

When 10g of 90% pure limestone is heated completely, the volume (in liters) of $C{O}_2$ liberated at STP?

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

Answer 1

Answer: (A)

$2.016$ L

$CaC{O}_3\stackrel{△}{\to }CaO+C{O}_2$

Ratio of moles: $1$ $:$ $1$

$10g$ of $90$% pure limestone $=9gCaC{O}_3$

Molar Weight of $CaC{O}_3=100g/mol$

Number of moles of $CaC{O}_3$ heated $=\frac{9}{100}moles$

$\frac{9}{100}molesCaC{O}_3:\frac{9}{100}molesC{O}_2$ will be liberated.

$1moleC{O}_2\to 22.4L$ at STP

$\frac{9}{100}molesC{O}_2\to xL$ at STP

$x=\frac{9}{100}\times 22.4=2.016L$