Question

What is the volume of $0.1N-HCl$ required to react completely with $1.0g$ of pure calcium carbonate?

• A. $100c{m}^3$
• B. $150c{m}^3$
• C. $250c{m}^3$
• D. $200c{m}^3$

Answer 1

Answer: (D)

Use the following formula to find $V$ of $HCl$ used
$V=\frac{w\times 1000}{\text{ equivalent weight }\times N}$
Given, $w=1.0gN=0.1$
$\underset{40+16\times 3+12=100}{CaC{O}_3}+2HCl\to CaC{l}_2+H_{2}O$
$\therefore 1$ mole of $CaC{O}_3$ reacts with $2$ moles of $HCl$
$\therefore$ Equivalent weight $=\frac{100}{2}=50$
$\therefore V=\frac{1\times 1000}{50\times 0.1}$
$\approx 200c{m}^3$