Question

The amount of calcium carbonate that reacts with $500cc$ of $0.5N$ hydrochloric acid is

• A. 50 g
• B. 25 g
• C. 12.5 g
• D. 100 g

Answer 1

Answer: (B)

Mass of $HCl$ in $500$ cc of $0.5NHCI$
$w=\frac{0.5\times 36.5\times 500}{1000}\left[\because N=\frac{w\times 1000}{eq.wt.xV(\text{ incc })}\right]$
$=9.125g$
The balanced equation for the reaction is
$\underset{100g}{\mathrm{Ca}}C{O}_3+\underset{2\times 36.5g}{2HCl}⟶CaC{l}_2+H_{2}O+C{O}_2$
$2\times 36.5g$ of $HCI$ react with $100g$ of $CaC{O}_3$
$\therefore 9.125g$ of $HCI$ will react with
$\frac{100\times 9.125}{2\times 36.5}=12.5g$ to $CaC{O}_3$