The mass of `CaCO_(3)` required to react completely with 20 mL of 1.0 M HCl as per the reaction:
`CaCO_(3) + 2HCl to CaCl_(2) + CO_(2) + H_(2)O` is (At. mass: Ca = 40, C = 12, O = 16)

Options

A. 1 g
B. 2 g
C. 10 g
D. 20 g of KOH in 200 mL of solution

Explanation

To find the mass of calcium carbonate (CaCO₃) required to react completely with 20 mL of 1.0 M HCl, we can follow these steps:

### Step 1: Calculate the number of moles of HCl
The formula to calculate the number of moles (n) is:
\[ n = \text{Molarity} \times \text{Volume (in liters)} \]

Given:
- Molarity of HCl = 1.0 M
- Volume of HCl = 20 mL = 20/1000 L = 0.020 L

Calculating the moles of HCl:
\[ n_{\text{HCl}} = 1.0 \, \text{mol/L} \times 0.020 \, \text{L} = 0.020 \, \text{mol} \]

### Step 2: Use the stoichiometry of the reaction
The balanced chemical equation is:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2O \]

From the equation, we see that 1 mole of CaCO₃ reacts with 2 moles of HCl. Therefore, the number of moles of CaCO₃ required can be calculated as:
\[ n_{\text{CaCO}_3} = \frac{n_{\text{HCl}}}{2} = \frac{0.020 \, \text{mol}}{2} = 0.010 \, \text{mol} \]

### Step 3: Calculate the molar mass of CaCO₃
The molar mass of CaCO₃ can be calculated using the atomic masses:
- Ca = 40 g/mol
- C = 12 g/mol
- O = 16 g/mol (there are 3 O atoms)

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

### Step 4: Calculate the mass of CaCO₃ required
Using the number of moles and the molar mass, we can find the mass (W) using the formula:
\[ W = n \times \text{Molar mass} \]

Calculating the mass of CaCO₃:
\[ W = 0.010 \, \text{mol} \times 100 \, \text{g/mol} = 1.0 \, \text{g} \]

### Final Answer
The mass of CaCO₃ required to react completely with 20 mL of 1.0 M HCl is **1.0 g**.

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