Calculate the mass of 95% pure $CaCO_3$ that will be required to neutralize 50 mL of 0.5 M HCl solution accord

Question

Calculate the mass of 95% pure $CaCO_3$ that will be required to neutralize 50 mL of 0.5 M HCl solution according to the following reaction:
$$ CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + 2H_2O(l) $$
[Calculate up to the second place of decimal point]

Accepted Answer

1. **Calculate Moles of HCl:**
   Using the molarity formula [Class 12 Chemistry, Ch 1, Eq. 1.8]:
   $$ 	ext{Moles of HCl} = 	ext{Molarity} 	imes 	ext{Volume (L)} $$
   $$ n_{HCl} = 0.5 	ext{ mol L}^{-1} 	imes 0.050 	ext{ L} = 0.025 	ext{ mol} $$
2. **Determine Stoichiometry:**
   From the balanced equation, $1 	ext{ mol } CaCO_3$ reacts with $2 	ext{ mol } HCl$.
   $$ 	ext{Moles of pure } CaCO_3 = \frac{1}{2} 	imes n_{HCl} = \frac{0.025}{2} = 0.0125 	ext{ mol} $$
3. **Calculate Mass of Pure $CaCO_3$:**
   Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 	ext{ g mol}^{-1}$.
   $$ 	ext{Mass}_{pure} = 0.0125 	ext{ mol} 	imes 100 	ext{ g mol}^{-1} = 1.25 	ext{ g} $$
4. **Adjust for Purity:**
   The sample is 95% pure, meaning $0.95 	imes 	ext{Mass}_{sample} = 	ext{Mass}_{pure}$.
   $$ 	ext{Mass}_{sample} = \frac{1.25 	ext{ g}}{0.95} \approx 1.3157 	ext{ g} $$
   Rounding to two decimal places, the required mass is **1.32 g**.

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