`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. (a)`5.5 g `
B. (b)` 11 g`
C. (c )`22 g `
D. (d)` 33 g`

Explanation

To solve the problem of how much weight of \( CO_2 \) will be obtained from the reaction of \( 100 \, g \) of \( CaCO_3 \) 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:
\[ CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O \]
This equation shows that 1 mole of \( CaCO_3 \) reacts with 2 moles of \( HCl \) to produce 1 mole of \( CO_2 \).

### Step 2: Calculate the number of moles of \( CaCO_3 \)
To find the number of moles of \( CaCO_3 \), we use its 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 × 3 = 48 g/mol

Thus, the molar mass of \( CaCO_3 \) is:
\[ 40 + 12 + 48 = 100 \, g/mol \]
Now, we can calculate the number of moles of \( CaCO_3 \):
\[ \text{Number of moles of } CaCO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, g}{100 \, g/mol} = 1 \, mol \]

### Step 3: Calculate the number of moles of \( HCl \)
Given that the normality of \( HCl \) is 1 N and the volume is 1 L, we can find the number of equivalents:
\[ \text{Number of equivalents of } HCl = \text{Normality} \times \text{Volume} = 1 \, N \times 1 \, L = 1 \, eq \]
Since \( HCl \) is a monoprotic acid, the number of moles of \( HCl \) is equal to the number of equivalents:
\[ \text{Number of moles of } HCl = 1 \, mol \]

### Step 4: Identify the limiting reagent
From the balanced equation, we see that 2 moles of \( HCl \) are required for every mole of \( CaCO_3 \). Since we have only 1 mole of \( HCl \) available, \( HCl \) is the limiting reagent.

### Step 5: Calculate the moles of \( CO_2 \) produced
According to the stoichiometry of the reaction:
\[ 2 \, moles \, HCl \rightarrow 1 \, mole \, CO_2 \]
Thus, if we have 1 mole of \( HCl \), the moles of \( CO_2 \) produced will be:
\[ \text{Moles of } CO_2 = \frac{1 \, mole \, HCl}{2} = 0.5 \, moles \]

### Step 6: Calculate the mass of \( CO_2 \)
The molar mass of \( CO_2 \) is:
\[ \text{C} = 12 \, g/mol + 2 \times \text{O} = 16 \, g/mol \times 2 = 32 \, g/mol \Rightarrow 12 + 32 = 44 \, g/mol \]
Now, we can calculate the mass of \( CO_2 \) produced:
\[ \text{Mass of } CO_2 = \text{Number of moles} \times \text{Molar mass} = 0.5 \, moles \times 44 \, g/mol = 22 \, g \]

### Final Answer
The weight of \( CO_2 \) obtained from the reaction is:
\[ \boxed{22 \, g} \]