To calculate the number of moles of calcium chloride (CaCl₂) obtained from 50 grams of limestone (CaCO₃) in the presence of excess hydrogen chloride (HCl), we need to follow a series of steps. Here's a detailed, step-by-step solution:
### Step 1: Determine the Molar Mass of Calcium Carbonate (CaCO₃)
We are given the atomic masses of:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol
Calcium carbonate (CaCO₃) consists of:
- 1 atom of Calcium
- 1 atom of Carbon
- 3 atoms of Oxygen
Thus, the molar mass of CaCO₃ is calculated as:
[tex]\[
\text{Molar mass of } CaCO_3 = 40 \, (\text{for Ca}) + 12 \, (\text{for C}) + 3 \times 16 \, (\text{for O}) = 100 \, \text{g/mol}
\][/tex]
### Step 2: Calculate the Moles of Calcium Carbonate (CaCO₃)
We have 50 grams of limestone (CaCO₃). To find the number of moles of CaCO₃, we use the formula:
[tex]\[
\text{Moles of } CaCO_3 = \frac{\text{mass of } CaCO_3}{\text{molar mass of } CaCO_3}
\][/tex]
[tex]\[
\text{Moles of } CaCO_3 = \frac{50 \, \text{g}}{100 \, \text{g/mol}} = 0.5 \, \text{moles}
\][/tex]
### Step 3: Determine the Relation in the Chemical Reaction
The balanced chemical equation for the reaction between CaCO₃ and HCl is:
[tex]\[
CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O
\][/tex]
From the equation, we can see that:
- 1 mole of CaCO₃ reacts to produce 1 mole of CaCl₂
Thus, the number of moles of CaCl₂ produced is directly equal to the number of moles of CaCO₃ reacted.
### Step 4: Calculate the Moles of Calcium Chloride (CaCl₂)
Given that we have 0.5 moles of CaCO₃, the number of moles of CaCl₂ produced will be:
[tex]\[
\text{Moles of } CaCl_2 = 0.5 \, \text{moles}
\][/tex]
### Conclusion
So, the number of moles of calcium chloride (CaCl₂) that can be obtained from 50 grams of limestone (CaCO₃) in the presence of excess hydrogen chloride (HCl) is:
[tex]\[
0.5 \, \text{moles}
\][/tex]
