Question

$CaC{O}_3+2HCl\to CaC{l}_2+H_{2}O+C{O}_2$

The volume of CO${}_2$ gas formed when 2.5 g calcium carbonate is dissolved in excess hydrochloric acid at 0${}^o$C and 1 atm pressure is:

[1 mole of any gas at 0${}^o$C and 1 atm pressure occupies 22.414 L volume].

• A. 1.12 L
• B. 56.0 L
• C. 0.28 L
• D. 0.56 L

Answer 1

Answer: (D)

0.56 L

$CaC{O}_3+2HCl\to CaC{l}_2+H_{2}O+C{O}_2$

100 g 44 g

1 mole 1 mole

22.4 L 22.4 L

We know that $1$ mole of any gas at $0^{o}C$ and $1atm$ pressure occupies $22.4L$ volume.

So, 100 g $CaC{O}_3$ forms $22.4L$ of $C{O}_2$

Hence, 2.5 g $CaC{O}_3$ will form $=\frac{2.5\times 22.4}{100}$

$=\frac{56}{100}L$ of $C{O}_2$

$=0.56L$ of $C{O}_2$

Hence, the correct option is $\text{D}$