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

Explanation

To find the mass of \( \text{CaCl}_2 \) formed from \( 10 \, \text{g} \) of \( \text{CaCO}_3 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation is:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Calculate the molar mass of \( \text{CaCO}_3 \)
- Calcium (Ca): \( 40 \, \text{g/mol} \)
- Carbon (C): \( 12 \, \text{g/mol} \)
- Oxygen (O): \( 16 \, \text{g/mol} \) (3 O atoms)

Calculating the molar mass:
\[ \text{Molar mass of } \text{CaCO}_3 = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \, \text{g/mol} \]

### Step 3: Calculate the molar mass of \( \text{CaCl}_2 \)
- Calcium (Ca): \( 40 \, \text{g/mol} \)
- Chlorine (Cl): \( 35.5 \, \text{g/mol} \) (2 Cl atoms)

Calculating the molar mass:
\[ \text{Molar mass of } \text{CaCl}_2 = 40 + (2 \times 35.5) = 40 + 71 = 111 \, \text{g/mol} \]

### Step 4: Use stoichiometry to find the mass of \( \text{CaCl}_2 \) produced
From the balanced equation, we see that:
1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CaCl}_2 \).

Now, we need to find out how many moles of \( \text{CaCO}_3 \) are in \( 10 \, \text{g} \):
\[ \text{Moles of } \text{CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \, \text{g}}{100 \, \text{g/mol}} = 0.1 \, \text{mol} \]

Since the ratio of \( \text{CaCO}_3 \) to \( \text{CaCl}_2 \) is 1:1, \( 0.1 \, \text{mol} \) of \( \text{CaCO}_3 \) will produce \( 0.1 \, \text{mol} \) of \( \text{CaCl}_2 \).

### Step 5: Calculate the mass of \( \text{CaCl}_2 \) produced
Now we can calculate the mass of \( \text{CaCl}_2 \) produced:
\[ \text{Mass of } \text{CaCl}_2 = \text{moles} \times \text{molar mass} = 0.1 \, \text{mol} \times 111 \, \text{g/mol} = 11.1 \, \text{g} \]

### Final Answer
The mass of \( \text{CaCl}_2 \) formed from \( 10 \, \text{g} \) of \( \text{CaCO}_3 \) is \( 11.1 \, \text{g} \).

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