`100 g CaCO_(3)` reacts with `1 litre 1 N HCl`. On completion of reaction how much weight of `CO_(2)` will be obtain

Options

A. 55g
B. 11g
C. 22g
D. 44g

Explanation

To solve the problem of how much weight of \( CO_2 \) will be obtained when \( 100 \, g \) of \( CaCO_3 \) reacts with \( 1 \, L \) of \( 1 \, N \, HCl \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between calcium carbonate (\( CaCO_3 \)) and hydrochloric acid (\( HCl \)) can be represented as:
\[ CaCO_3 + 2HCl \rightarrow CO_2 + CaCl_2 + H_2O \]

### Step 2: Calculate the number of moles of \( CaCO_3 \)
To find the number of moles of \( CaCO_3 \), we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
The molar mass of \( CaCO_3 \) is calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (for 3 O, it is \( 16 \times 3 = 48 \, g/mol \))

Thus, the molar mass of \( CaCO_3 \) is:
\[ 40 + 12 + 48 = 100 \, g/mol \]

Now, substituting the values:
\[ \text{Number of moles of } CaCO_3 = \frac{100 \, g}{100 \, g/mol} = 1 \, mol \]

### Step 3: Calculate the number of moles of \( HCl \)
Given that the solution is \( 1 \, N \) and \( 1 \, L \), the number of moles of \( HCl \) can be calculated as:
\[ \text{Number of moles of } HCl = \text{Normality} \times \text{Volume (L)} = 1 \, N \times 1 \, L = 1 \, mol \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that:
- 2 moles of \( HCl \) react with 1 mole of \( CaCO_3 \).
- Therefore, 1 mole of \( HCl \) will react with \( \frac{1}{2} \) mole of \( CaCO_3 \).

Since we have 1 mole of \( CaCO_3 \) and only 1 mole of \( HCl \), \( HCl \) is the limiting reagent.

### Step 5: Calculate the amount of \( CO_2 \) produced
According to the reaction stoichiometry:
- 2 moles of \( HCl \) produce 1 mole of \( CO_2 \).
- Therefore, 1 mole of \( HCl \) will produce \( \frac{1}{2} \) mole of \( CO_2 \).

### Step 6: Calculate the mass of \( CO_2 \)
The molar mass of \( CO_2 \) is calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (for 2 O, it is \( 16 \times 2 = 32 \, g/mol \))

Thus, the molar mass of \( CO_2 \) is:
\[ 12 + 32 = 44 \, g/mol \]

Now, the mass of \( \frac{1}{2} \) mole of \( CO_2 \) is:
\[ \text{Mass of } CO_2 = \frac{1}{2} \times 44 \, g/mol = 22 \, g \]

### Final Answer
The weight of \( CO_2 \) obtained is \( 22 \, g \).

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