Question

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

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

Answer 1

The correct option is A $22.4L$

$\underset{\left(Limestone\right)}{\underset{}{CaC{O}_3}}⟶CaO+C{O}_2$

$1$ $mole$ of $CaC{O}_3$ would give $1$ $mole$ of $CaO$ and $C{O}_2$.

Now, $90\%$ of limestone is heater.

$\therefore \frac{90}{100}$ $mole$ of $CaC{O}_3\to 100\times \frac{90}{100}=90$ $gram$

Moles of $CaC{O}_3=\frac{90g}{90g}=1$ mole

$\therefore$ Moles of $C{O}_2=1$ mole

At STP,

$1$ $mole$ of gas $=22.4$ $L$

$\therefore$ Volume of $C{O}_2$ liberated $=22.4$ $L$