What is the volume (in liters) of ${\mathrm{CO}}_2$ liberated at STP, when $2.12\mathrm{gm}s$ of sodium carbonate $(\mathrm{MW}=106)$ is treated with excess dilute $\mathrm{HCl}$ ?

• Sodium carbonate when reacts with dilute hydrochloric acid to form sodium chloride, carbon dioxide and water.

           ${\mathrm{Na}}_2{\mathrm{CO}}_3+2\mathrm{HCl}\to 2\mathrm{NaCl}+{\mathrm{CO}}_2\uparrow +{\mathrm{H}}_2\mathrm{O}$

No. of moles of ${\mathrm{Na}}_2{\mathrm{CO}}_3=2.12/106=2\times {10}^{-2}$

 1 mole of ${\mathrm{Na}}_2{\mathrm{CO}}_3=1$ mole of ${\mathrm{CO}}_2$, which is $22.4 \mathrm{L}$ at STP.

So, $2\times {10}^{-2}$ moles of ${\mathrm{Na}}_2{\mathrm{CO}}_3=\mathrm{x}$ mole of ${\mathrm{CO}}_2$

$\Rightarrow \mathrm{x}=22.4\times 2\times {10}^{-2}=0.448\mathrm{L}$

So, the volume of ${\mathrm{CO}}_2$ that is liberated at STP is $0.448 \mathrm{L}$.