Question
Calculate the number of moler of cacl2 that can be abatined from 25g of limestone (caco3) in the gresence of excess acid (Hcl )
Answers
GPT-4o mini
To determine the number of moles of calcium chloride (CaCl₂) that can be obtained from 25 grams of limestone (calcium carbonate, CaCO₃) in the presence of excess hydrochloric acid (HCl), we can follow these steps:
1. **Write the balanced chemical reactions**:
- The reaction of calcium carbonate (CaCO₃) with hydrochloric acid (HCl) is as follows:
\[
\text{CaCO}_3 (s) + 2\text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g)
\]
From this equation, we see that 1 mole of CaCO₃ produces 1 mole of CaCl₂.
2. **Calculate the molar mass of CaCO₃**:
- The molar mass of calcium carbonate (CaCO₃) is calculated by adding the molar masses of each element:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
\[
\text{Molar mass of CaCO}_3 = 40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol}
\]
3. **Calculate the number of moles of CaCO₃ in 25 grams**:
\[
\text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{25 \text{ g}}{100.09 \text{ g/mol}} \approx 0.2498 \text{ moles}
\]
4. **Determine the number of moles of CaCl₂ produced**:
According to the balanced equation, 1 mole of CaCO₃ produces 1 mole of CaCl₂. Therefore, the moles of CaCl₂ produced will be the same as the moles of CaCO₃ reacted.
\[
\text{Moles of CaCl}_2 = \text{Moles of CaCO}_3 = 0.2498 \text{ moles}
\]
Therefore, from 25 grams of limestone (CaCO₃), approximately 0.2498 moles of calcium chloride (CaCl₂) can be obtained in the presence of excess hydrochloric acid (HCl).
1. **Write the balanced chemical reactions**:
- The reaction of calcium carbonate (CaCO₃) with hydrochloric acid (HCl) is as follows:
\[
\text{CaCO}_3 (s) + 2\text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g)
\]
From this equation, we see that 1 mole of CaCO₃ produces 1 mole of CaCl₂.
2. **Calculate the molar mass of CaCO₃**:
- The molar mass of calcium carbonate (CaCO₃) is calculated by adding the molar masses of each element:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
\[
\text{Molar mass of CaCO}_3 = 40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol}
\]
3. **Calculate the number of moles of CaCO₃ in 25 grams**:
\[
\text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{25 \text{ g}}{100.09 \text{ g/mol}} \approx 0.2498 \text{ moles}
\]
4. **Determine the number of moles of CaCl₂ produced**:
According to the balanced equation, 1 mole of CaCO₃ produces 1 mole of CaCl₂. Therefore, the moles of CaCl₂ produced will be the same as the moles of CaCO₃ reacted.
\[
\text{Moles of CaCl}_2 = \text{Moles of CaCO}_3 = 0.2498 \text{ moles}
\]
Therefore, from 25 grams of limestone (CaCO₃), approximately 0.2498 moles of calcium chloride (CaCl₂) can be obtained in the presence of excess hydrochloric acid (HCl).