Question

100 mL of N/5 HCl was added to 1 g of pure ${CaCO_3}$ What would remain after the reaction?

• A. 0.5 g of of $CaCO_3$
• B. Neither $CaCO_3$ nor HCI
• C. 50 mL of HCI
• D. 25 mL of HCI

Answer 1

Answer: (B)

Neither $CaCO_3$ nor HCI

Answer 2

No. of moles of $CaCO_3 = \frac{w}{M} = \frac{1}{100}$
= 0.01 mol
[$\because$ Molar mass of $CaCO_3 = 100\, g\, mol^{-1}$]
100 mL of N/5 HCl = $\frac{100}{1000} \times \frac{1}{5}$
= 0.02 mol
The reaction is represented as :
${ CaCO_3 + 2HCl -> CaCl_2 + CO_2 + H_2 O}$
1 mole of $CaCO_3$ reacts with 2 moles of HCI. Hence, 0.01 mole of $CaCO_3$ will react completely with 0.02 mole of HCI.