`CaCO_(3)` +` 2HCl to CaCl_(2)` + `CO_(2)`+ `H_(2)O`
what will be the amount of `CaCl_(2)` when 10 g` CaCO_(3)` and
200 mL of 0.75 M HCI is used in the reaction ?

Options

A. 83.25 g
B. 16.65 g
C. 11.1 g
D. 8.325 g

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of HCl
We are given the volume of HCl (200 mL) and its molarity (0.75 M). We can calculate the number of moles using the formula:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]

Convert the volume from mL to L:

\[ 200 \, \text{mL} = \frac{200}{1000} \, \text{L} = 0.2 \, \text{L} \]

Now, calculate the number of moles of HCl:

\[ \text{Number of moles of HCl} = 0.75 \, \text{mol/L} \times 0.2 \, \text{L} = 0.15 \, \text{moles} \]

### Step 2: Calculate the number of moles of CaCO₃
We are given the mass of CaCO₃ (10 g) and need to find its number of moles. The molar mass of CaCO₃ is approximately 100 g/mol.

Using the formula:

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

Calculate the number of moles of CaCO₃:

\[ \text{Number of moles of CaCO₃} = \frac{10 \, \text{g}}{100 \, \text{g/mol}} = 0.1 \, \text{moles} \]

### Step 3: Identify the limiting reagent
From the balanced chemical equation:

\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2O \]

We see that 1 mole of CaCO₃ reacts with 2 moles of HCl.

- We have 0.1 moles of CaCO₃, which would require:
\[ 0.1 \, \text{moles of CaCO}_3 \times 2 = 0.2 \, \text{moles of HCl} \]

- We have only 0.15 moles of HCl available.

Since we have less HCl than required, HCl is the limiting reagent.

### Step 4: Calculate the number of moles of CaCl₂ produced
From the balanced equation, 2 moles of HCl produce 1 mole of CaCl₂. Therefore, the moles of CaCl₂ produced from 0.15 moles of HCl is:

\[ \text{Moles of CaCl}_2 = \frac{0.15 \, \text{moles of HCl}}{2} = 0.075 \, \text{moles} \]

### Step 5: Calculate the mass of CaCl₂ produced
Now, we need to find the mass of CaCl₂ produced. The molar mass of CaCl₂ is approximately 111 g/mol.

Using the formula:

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

Calculate the mass of CaCl₂:

\[ \text{Mass of CaCl}_2 = 0.075 \, \text{moles} \times 111 \, \text{g/mol} = 8.325 \, \text{g} \]

### Final Answer
The amount of CaCl₂ produced is **8.325 grams**.

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