With the help of the equation `CaCO_(3) + 2HCl to CaCl_(2) + H_(2)O + CO_(2)`, calculate the following :
The volume of `CO_(2)` formed at STP from 10 g `CaCO_(3)`.

Explanation

To solve the problem of calculating the volume of \( CO_2 \) formed at STP from 10 g of \( CaCO_3 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \]

### Step 2: Calculate the molar mass of \( CaCO_3 \)
The molar mass of \( CaCO_3 \) (Calcium Carbonate) can be calculated as follows:
- Calcium (Ca) = 40.08 g/mol
- Carbon (C) = 12.01 g/mol
- Oxygen (O) = 16.00 g/mol (3 O atoms)

\[ \text{Molar mass of } CaCO_3 = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \text{ g/mol} \]

### Step 3: Calculate the number of moles of \( CaCO_3 \)
Using the mass of \( CaCO_3 \) provided (10 g), we can calculate the number of moles:
\[ \text{Number of moles of } CaCO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{100.09 \text{ g/mol}} \approx 0.0999 \text{ moles} \]

### Step 4: Use the stoichiometry of the reaction
From the balanced equation, we see that 1 mole of \( CaCO_3 \) produces 1 mole of \( CO_2 \). Therefore, the number of moles of \( CO_2 \) produced will be equal to the number of moles of \( CaCO_3 \):
\[ \text{Moles of } CO_2 = 0.0999 \text{ moles} \]

### Step 5: Calculate the volume of \( CO_2 \) at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus, the volume of \( CO_2 \) produced can be calculated as follows:
\[ \text{Volume of } CO_2 = \text{moles} \times 22.4 \text{ L/mol} = 0.0999 \text{ moles} \times 22.4 \text{ L/mol} \approx 2.24 \text{ L} \]

### Final Answer
The volume of \( CO_2 \) formed at STP from 10 g of \( CaCO_3 \) is approximately **2.24 liters**.

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