2.5 g of `CaCO_(3)` was placed in 50 ml of a solution of HCl.1.05 g of `CaCO_(3)` was left after the reaction. Calculate:
the Molarity of HCl

Answer

To calculate the molarity 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 as:
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{CO}_3 (aq) \]

### Step 2: Calculate the amount of CaCO₃ that reacted
Initially, we had 2.5 g of CaCO₃, and after the reaction, we have 1.05 g left. Therefore, the amount of CaCO₃ that reacted is:
\[ \text{Amount reacted} = \text{Initial amount} - \text{Remaining amount} \]
\[ \text{Amount reacted} = 2.5 \, \text{g} - 1.05 \, \text{g} = 1.45 \, \text{g} \]

### Step 3: Convert grams of CaCO₃ to moles
To find the number of moles of CaCO₃ that reacted, we use its molar mass. The molar mass of CaCO₃ is approximately 100 g/mol.
\[ \text{Moles of CaCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.45 \, \text{g}}{100 \, \text{g/mol}} = 0.0145 \, \text{mol} \]

### Step 4: Use the stoichiometry of the reaction to find moles of HCl
From the balanced equation, we see that 1 mole of CaCO₃ reacts with 2 moles of HCl. Therefore, the moles of HCl that reacted can be calculated as:
\[ \text{Moles of HCl} = 2 \times \text{Moles of CaCO}_3 \]
\[ \text{Moles of HCl} = 2 \times 0.0145 \, \text{mol} = 0.029 \, \text{mol} \]

### Step 5: Calculate the molarity of HCl
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters. The volume of HCl solution is given as 50 mL, which is equivalent to 0.050 L.
\[ \text{Molarity} = \frac{\text{Moles of HCl}}{\text{Volume of solution in L}} \]
\[ \text{Molarity} = \frac{0.029 \, \text{mol}}{0.050 \, \text{L}} = 0.58 \, \text{M} \]

### Final Answer
The molarity of the HCl solution is **0.58 M**.

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