Question

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

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

Answer 1

Answer: (B)

25 g

Answer 2

Mass of HCl in 500 cc of 0.5 N HCI
$w = \frac{0.5 \times 36.5 \times 500}{1000} \left[ \because \:\:\: N = \frac{w \times 1000}{e wt.x V (in cc)}\right]$
$= 9.125 \, g$
The balanced equation for the reaction is
{\underset{\text{100 g}}{ {Ca}}C0_3$+$\underset{\text{2 $\times$ 36.5 g}}{{2HCl}} -> CaCl_2 + H_2O + CO_2}
${2 \times 36.5\, g}$ of HCI react with 100 g of $CaCO_3$
$\therefore$ 9.125 g of HCI will react with
$\frac{100 \times 9.125}{2 \times 36.5} = 12.5 \, g$ to $CaCO_3$