Question

How many grams of calcium carbonate ($\mathrm{CaCO}_3$) are required to completely neutralize 200 mL of 0.5 M HCl? Given the reaction: $$ \mathrm{CaCO}_3 + 2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2 + \mathrm{H}_2 \mathrm{O} + \mathrm{CO}_2 $$ (Molar mass of $\mathrm{CaCO}_3 = 100 \, \text{g/mol}$)

• A. 5.0 g
• B. 10.0 g
• C. 2.5 g
• D. 7.5 g

Hint:

In neutralization reactions, use the stoichiometric ratio from the balanced equation to relate moles of reactants, then convert to mass using molar mass.

Answer 1

The Correct Option is A

Given: \[ \text{Volume of HCl} = 200 \, \text{mL} = 0.2 \, \text{L}, \quad \text{Molarity of HCl} = 0.5 \, \text{M}, \quad \text{Molar mass of } \mathrm{CaCO}_3 = 100 \, \text{g/mol} \] Step 1: Calculate Moles of HCl
Moles of HCl = Molarity \(\times\) Volume (in liters): \[ \text{Moles of HCl} = 0.5 \cdot 0.2 = 0.1 \, \text{mol} \] Step 2: Use Stoichiometry
From the reaction, 1 mole of \(\mathrm{CaCO}_3\) neutralizes 2 moles of HCl. Thus, moles of \(\mathrm{CaCO}_3\) required: \[ \text{Moles of } \mathrm{CaCO}_3 = \frac{\text{Moles of HCl}}{2} = \frac{0.1}{2} = 0.05 \, \text{mol} \] Step 3: Calculate Mass of \(\mathrm{CaCO}_3\)
Mass = Moles \(\times\) Molar mass: \[ \text{Mass of } \mathrm{CaCO}_3 = 0.05 \cdot 100 = 5.0 \, \text{g} \] Thus, the mass required is: \[ \boxed{5.0 \, \text{g}} \]