Question

A certain weight of pure caco3 is made to react completely with 200 ml of a hcl solution to give 224 ml of co2 gas at stp. the normality of the hcl solution is:

Answer 1

Equation for the reaction :

2 HCl (aq) + CaCO₃ (s) —> CaCl₂ (aq) + H₂O(l) + CO₂ (g)

The mole ratio is 2 : 1

Moles of CO₂ :

224 / 22400 = 0.01 moles

Moles of HCl :

0.01 × 2 = 0.02 moles

Molarity of HCl :

(1000 / 200) × 0.02 = 0.1

Normality = Molarity × Equivalents (number of hydrogen ions in the acid)

Equivalents = 1

N = 1 × 0.1 = 0.1 n

Answer 2

Answer: The normality of the HCl solution is 0.1 N.

Explanation:

$CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O$

Volume occupied by the carbon-dioxide gas = 224 ml= 0.224 L

At STP, the 1 mol of gases occupy 22.4 L of the volume.

Then 0.224 l of the volume will be occupied by :

$\frac{1}{22.4 L}\times 0.224 L=0.01 mol$ of $CO_2$ gas

According to reaction, 1 mole of $CO_2$ gas are produced by 2 moles of HCl , then 0.01 mole of gas will be produced by;$\frac{2}{1}\times 0.01$ that is 0.02 moles of HCl.

Volume of HCl solution = 200 ml = 0.2 L

$Normality=\frac{\text{Number of moles of HCl}}{\text{gram equivalents of HCl}\times \text{Volume od HCl (L)}}$

$Normality=\frac{0.02}{1\times 0.2}=0.1 N$

The normality of the HCl solution is 0.1 N.