A 50.0 mL solution of Ca(OH)2 with an unknown concentration was titrated with 0.340 M HNO3. To reach the endpoint, a total of 28.4 mL of HNO, was required. Given that 0.00966 mol of HNO, are used in the titration, What quantity in moles of Ca(OH)2 had to be present in the initial reaction?
A 50.0 mL solution of Ca(OH)2 with an unknown concentration was titrated with 0.340 M HNO3. To reach the endpoint, a total of 28.4 mL of HNO, was required. Given that 0.00966 mol of HNO, are used in the titration, What quantity in moles of Ca(OH)2 had to be present in the initial reaction?
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![### Titration Problem:
A 50.0 mL solution of Ca(OH)₂ with an unknown concentration was titrated with 0.340 M HNO₃. To reach the endpoint, a total of 28.4 mL of HNO₃ was required.
Given that 0.00966 mol of HNO₃ are used in the titration, what quantity in moles of Ca(OH)₂ had to be present in the initial reaction?
### Explanation:
To solve this problem, consider the balanced chemical equation between Ca(OH)₂ and HNO₃:
\[ \text{Ca(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Ca(NO}_3)_2 + 2\text{H}_2\text{O} \]
This indicates that 2 moles of HNO₃ react with 1 mole of Ca(OH)₂. You can use the stoichiometric relationship between the reactants to find the initial moles of Ca(OH)₂.
### Steps to Solve:
1. **Identify the Moles of HNO₃:**
- Provided that 0.00966 mol of HNO₃ were used in the reaction.
2. **Stoichiometric Conversion:**
- Since 2 moles of HNO₃ react with 1 mole of Ca(OH)₂, we can calculate the moles of Ca(OH)₂ needed initially by dividing the moles of HNO₃ by 2.
3. **Calculation:**
- Moles of Ca(OH)₂ = \( \frac{0.00966 \text{ mol HNO}_3}{2} = 0.00483 \text{ mol Ca(OH)}_2 \)
The initial reaction required 0.00483 moles of Ca(OH)₂.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F458e827c-2686-4898-b7e9-4e2a75f62872%2Fb24d3d11-fa81-42d1-814b-2c57fb705b87%2Faej0ln9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Titration Problem:
A 50.0 mL solution of Ca(OH)₂ with an unknown concentration was titrated with 0.340 M HNO₃. To reach the endpoint, a total of 28.4 mL of HNO₃ was required.
Given that 0.00966 mol of HNO₃ are used in the titration, what quantity in moles of Ca(OH)₂ had to be present in the initial reaction?
### Explanation:
To solve this problem, consider the balanced chemical equation between Ca(OH)₂ and HNO₃:
\[ \text{Ca(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Ca(NO}_3)_2 + 2\text{H}_2\text{O} \]
This indicates that 2 moles of HNO₃ react with 1 mole of Ca(OH)₂. You can use the stoichiometric relationship between the reactants to find the initial moles of Ca(OH)₂.
### Steps to Solve:
1. **Identify the Moles of HNO₃:**
- Provided that 0.00966 mol of HNO₃ were used in the reaction.
2. **Stoichiometric Conversion:**
- Since 2 moles of HNO₃ react with 1 mole of Ca(OH)₂, we can calculate the moles of Ca(OH)₂ needed initially by dividing the moles of HNO₃ by 2.
3. **Calculation:**
- Moles of Ca(OH)₂ = \( \frac{0.00966 \text{ mol HNO}_3}{2} = 0.00483 \text{ mol Ca(OH)}_2 \)
The initial reaction required 0.00483 moles of Ca(OH)₂.
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