What mass of `CaCO_(3)` is required to react completely with 25 ml of `0.75 MHCI`?

Options

A. 1 g
B. 0.3 g
C. 0.8 g
D. 0.93 g

Answer

To find the mass of calcium carbonate (CaCO₃) required to react completely with 25 mL of 0.75 M HCl, we can follow these steps:

### Step 1: Calculate the number of moles of HCl
We know that molarity (M) is defined as the number of moles of solute (n) divided by the volume of solution in liters (V).

\[ \text{Molarity (M)} = \frac{\text{Number of moles (n)}}{\text{Volume (L)}} \]

Given:
- Molarity of HCl = 0.75 M
- Volume of HCl = 25 mL = 0.025 L

Using the formula, we can rearrange it to find the number of moles of HCl:

\[ n = M \times V \]

Substituting the values:

\[ n = 0.75 \, \text{mol/L} \times 0.025 \, \text{L} = 0.01875 \, \text{mol} \]

### Step 2: Write the balanced chemical equation
The balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is:

\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]

From the equation, we see that 1 mole of CaCO₃ reacts with 2 moles of HCl.

### Step 3: Calculate the number of moles of CaCO₃ required
From the balanced equation, we can determine the moles of CaCO₃ needed to react with the moles of HCl:

\[ \text{Moles of CaCO}_3 = \frac{\text{Moles of HCl}}{2} = \frac{0.01875}{2} = 0.009375 \, \text{mol} \]

### Step 4: Calculate the mass of CaCO₃ required
To find the mass of CaCO₃, we use the formula:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

The molar mass of CaCO₃ is approximately:

- Ca: 40.08 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol × 3 = 48.00 g/mol

Total molar mass of CaCO₃ = 40.08 + 12.01 + 48.00 = 100.09 g/mol

Now, substituting the values:

\[ \text{Mass of CaCO}_3 = 0.009375 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 0.937 \, \text{g} \]

### Conclusion
The mass of CaCO₃ required to react completely with 25 mL of 0.75 M HCl is approximately **0.937 g**.

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