A small amount of `CaCO_(3)` completely neutralises 525 mL of 0.1 N HCl and no acid is left in the end. After converting all calcium chlorine to `CaSO_(4)`, how much plaster of Paris can be obtained ?

Options

A. 3.8 gm
B. 14.2 gm
C. 5.6 gm
D. 17.43 gm

Answer

To solve the problem step by step, we will follow the chemical reactions involved and the calculations needed to find out how much plaster of Paris can be obtained after neutralizing the hydrochloric acid with calcium carbonate.

### Step 1: Determine the amount of HCl neutralized
We know that:
- Volume of HCl = 525 mL
- Normality of HCl = 0.1 N

To find the equivalents of HCl, we use the formula:
\[ \text{Equivalents of HCl} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Equivalents of HCl} = 0.1 \, \text{N} \times 0.525 \, \text{L} = 0.0525 \, \text{equivalents} \]

### Step 2: Write the neutralization reaction
The reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]
From the equation, we can see that 1 equivalent of CaCO₃ reacts with 2 equivalents of HCl.

### Step 3: Calculate the equivalents of CaCO₃
Since 0.0525 equivalents of HCl are neutralized, the equivalents of CaCO₃ can be calculated as:
\[ \text{Equivalents of CaCO}_3 = \frac{\text{Equivalents of HCl}}{2} = \frac{0.0525}{2} = 0.02625 \, \text{equivalents} \]

### Step 4: Convert CaCl₂ to CaSO₄
After the reaction, we convert calcium chloride (CaCl₂) to calcium sulfate (CaSO₄). The reaction is:
\[ \text{CaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2 \text{NaCl} \]
From this reaction, we can see that 1 equivalent of CaCl₂ gives 1 equivalent of CaSO₄.

Thus, the equivalents of CaSO₄ produced will also be 0.02625 equivalents.

### Step 5: Calculate the mass of plaster of Paris (CaSO₄·0.5H₂O)
The molar mass of plaster of Paris (CaSO₄·0.5H₂O) is:
- Ca = 40 g/mol
- S = 32 g/mol
- O₄ = 16 g/mol × 4 = 64 g/mol
- 0.5 H₂O = 9 g/mol (H₂O = 18 g/mol)

Total molar mass = 40 + 32 + 64 + 9 = 145 g/mol

The equivalent mass of plaster of Paris is:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{n} = \frac{145 \, \text{g/mol}}{2} = 72.5 \, \text{g/equiv} \]

### Step 6: Calculate the mass of plaster of Paris produced
Using the formula:
\[ \text{Mass of plaster of Paris} = \text{Equivalents} \times \text{Equivalent mass} \]
\[ \text{Mass of plaster of Paris} = 0.02625 \, \text{equivalents} \times 72.5 \, \text{g/equiv} = 1.903125 \, \text{g} \]

### Conclusion
The amount of plaster of Paris that can be obtained is approximately 1.9 g.