If 20 g of `CaCO_(3)` is treated with 100 mL of 20% HCl solution, the amount of `CO_(2)` produced is

Options

A. `22.4 L`
B. `8.80 g`
C. `4.40 g`
D. `2.24 L`

Explanation

To solve the problem of how much CO₂ is produced when 20 g of CaCO₃ is treated with 100 mL of 20% HCl solution, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) can be represented as:
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2O (l) \]

### Step 2: Calculate the number of moles of CaCO₃
To find the number of moles of CaCO₃, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of CaCO₃ 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 CaCO₃ = 40 + 12 + 48 = 100 g/mol.

Now, calculate the moles of CaCO₃:
\[ \text{Moles of CaCO}_3 = \frac{20 \, \text{g}}{100 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 3: Calculate the number of moles of HCl
The concentration of HCl is given as 20% w/v, which means there are 20 g of HCl in 100 mL of solution.

To find the number of moles of HCl, we first calculate the molar mass of HCl:
- Hydrogen (H): 1 g/mol
- Chlorine (Cl): 35.5 g/mol

Thus, the molar mass of HCl = 1 + 35.5 = 36.5 g/mol.

Now, calculate the moles of HCl:
\[ \text{Moles of HCl} = \frac{20 \, \text{g}}{36.5 \, \text{g/mol}} \approx 0.549 \, \text{mol} \]

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

For 0.2 moles of CaCO₃, the required moles of HCl are:
\[ 0.2 \, \text{mol CaCO}_3 \times 2 \, \text{mol HCl/mol CaCO}_3 = 0.4 \, \text{mol HCl} \]

Since we have approximately 0.549 moles of HCl available, which is more than enough to react with 0.2 moles of CaCO₃, CaCO₃ is the limiting reagent.

### Step 5: Calculate the amount of CO₂ produced
According to the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, 0.2 moles of CaCO₃ will produce:
\[ 0.2 \, \text{mol CO}_2 \]

### Step 6: Calculate the mass of CO₂ produced
To find the mass of CO₂ produced, we calculate the molar mass of CO₂:
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

Thus, the molar mass of CO₂ = 12 + 32 = 44 g/mol.

Now, calculate the mass of CO₂:
\[ \text{Mass of CO}_2 = \text{moles} \times \text{molar mass} = 0.2 \, \text{mol} \times 44 \, \text{g/mol} = 8.8 \, \text{g} \]

### Final Answer
The amount of CO₂ produced is **8.8 grams**.

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