Express of `CO_(2)` is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonated was completely neutralized with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At mass of carbon = 40)

Options

A. ` 200 cm^3`
B. `500 cm^3`
C. `400 cm^3`
D. `300 cm^3`

Answer

To solve the problem step by step, we will follow the chemical reactions and calculations involved in the process.

### Step 1: Write the Reaction
The reaction between calcium hydroxide (Ca(OH)₂) and carbon dioxide (CO₂) produces calcium carbonate (CaCO₃) and water (H₂O). The balanced equation is:

\[ \text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2O \]

### Step 2: Calculate the Moles of Calcium Hydroxide
We are given the volume and molarity of the calcium hydroxide solution.

- Volume of Ca(OH)₂ solution = 50 mL = 0.050 L
- Molarity of Ca(OH)₂ = 0.5 M

To find the number of moles of calcium hydroxide:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} = 0.5 \, \text{mol/L} \times 0.050 \, \text{L} = 0.025 \, \text{mol} \]

### Step 3: Calculate the Moles of Calcium Carbonate
From the balanced equation, we see that 1 mole of Ca(OH)₂ produces 1 mole of CaCO₃. Therefore, the number of moles of calcium carbonate produced is also:

\[ \text{Moles of CaCO}_3 = 0.025 \, \text{mol} \]

### Step 4: Convert Moles of Calcium Carbonate to Milliequivalents
The valency factor of calcium carbonate (CaCO₃) is 2 because it can donate 2 equivalents of Ca²⁺ ions. To find the number of milliequivalents:

\[ \text{Milliequivalents of CaCO}_3 = \text{Moles} \times \text{Valency factor} \times 1000 = 0.025 \, \text{mol} \times 2 \times 1000 = 50 \, \text{meq} \]

### Step 5: Calculate the Volume of Hydrochloric Acid Required
The neutralization reaction between calcium carbonate and hydrochloric acid (HCl) can be represented as:

\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2O + \text{CO}_2 \]

From the reaction, we see that 1 equivalent of CaCO₃ reacts with 2 equivalents of HCl. Therefore, the number of equivalents of HCl required is:

\[ \text{Equivalents of HCl} = \text{Milliequivalents of CaCO}_3 = 50 \, \text{meq} \]

Given that the normality (N) of HCl is 0.1 N, we can use the formula:

\[ \text{Normality} \times \text{Volume} = \text{Equivalents} \]

Let \( V \) be the volume of HCl in liters:

\[ 0.1 \, \text{N} \times V = 50 \, \text{meq} \quad \text{(or 0.050 eq)} \]

Thus,

\[ V = \frac{50 \, \text{meq}}{0.1 \, \text{N}} = \frac{50}{0.1} = 500 \, \text{mL} \]

### Final Answer
The volume of hydrochloric acid required is **500 mL**.

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