Question

What is the volume (in L) of carbon dioxide liberated at STP when $2.12$ grams of sodium carbonate (mol. wt. $=106$) is treated with excess dilute $HCl$?

• A. $2.28$
• B. $0.448$
• C. $44.8$
• D. $22.4$

Answer 1

The correct option is B $0.448$

The balanced chemical equation is, $N{a}_{2}C{O}_3+2HCl\to 2NaCl+C{O}_2+H_{2}O$.

$1$ mole of sodium carbonate ($106$ g) will produce $1$ mol ($22.4$ L at STP) of carbon dioxide.

Thus, $2.12$ g ($\frac{2.12}{106}$ mol) of sodium carbonate will produced $(\frac{2.12}{106}\times 22.4=0.448L)$ at STP of carbon dioxide.

So, the correct option is $B$