If 20.0 g of `CaCO_(3)` is treated with 20.0 g of HCl, how many grams of `CO_(2)` will be produced ?

Explanation

To solve the problem of how many grams of CO₂ will be produced when 20.0 g of CaCO₃ is treated with 20.0 g of HCl, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) can be represented by the following balanced equation:

\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \]

### Step 2: Calculate the molar masses
Next, we need to calculate the molar masses of the reactants and products involved in the reaction.

- **Molar mass of CaCO₃**:
- Calcium (Ca): 40.0 g/mol
- Carbon (C): 12.0 g/mol
- Oxygen (O): 16.0 g/mol × 3 = 48.0 g/mol
- Total: \( 40.0 + 12.0 + 48.0 = 100.0 \) g/mol

- **Molar mass of HCl**:
- Hydrogen (H): 1.0 g/mol
- Chlorine (Cl): 35.5 g/mol
- Total: \( 1.0 + 35.5 = 36.5 \) g/mol

- **Molar mass of CO₂**:
- Carbon (C): 12.0 g/mol
- Oxygen (O): 16.0 g/mol × 2 = 32.0 g/mol
- Total: \( 12.0 + 32.0 = 44.0 \) g/mol

### Step 3: Calculate moles of reactants
Now, we can calculate the number of moles of each reactant.

- **Moles of CaCO₃**:
\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{100.0 \, \text{g/mol}} = 0.200 \, \text{mol} \]

- **Moles of HCl**:
\[ \text{Moles of HCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{36.5 \, \text{g/mol}} \approx 0.549 \, \text{mol} \]

### Step 4: Determine the limiting reactant
From the balanced equation, we see that 1 mole of CaCO₃ reacts with 2 moles of HCl. Therefore, we need:

\[ \text{Required moles of HCl for 0.200 mol of CaCO}_3 = 0.200 \, \text{mol} \times 2 = 0.400 \, \text{mol} \]

Since we have 0.549 moles of HCl available, HCl is in excess, and CaCO₃ is the limiting reactant.

### Step 5: Calculate the amount of CO₂ produced
From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the moles of CO₂ produced will be equal to the moles of CaCO₃ reacted:

\[ \text{Moles of CO}_2 = \text{Moles of CaCO}_3 = 0.200 \, \text{mol} \]

Now, we can convert moles of CO₂ to grams:

\[ \text{Mass of CO}_2 = \text{moles} \times \text{molar mass} = 0.200 \, \text{mol} \times 44.0 \, \text{g/mol} = 8.8 \, \text{g} \]

### Final Answer
The mass of CO₂ produced is **8.8 g**.

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