How many moles of Agl will be formed when 75.0 mL of 0.300MAGNO3 is completely reacted according to the balanced chemical reaction: 2 AGNO3(aq) + Cal2(aq) → 2 Agl(s) + Ca(NO3)2(aq)
How many moles of Agl will be formed when 75.0 mL of 0.300MAGNO3 is completely reacted according to the balanced chemical reaction: 2 AGNO3(aq) + Cal2(aq) → 2 Agl(s) + Ca(NO3)2(aq)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Stoichiometry Calculation: How Many Moles of AgI will be Formed?**
**Problem Statement:**
How many moles of AgI will be formed when 75.0 mL of 0.300 M AgNO₃ is completely reacted according to the balanced chemical reaction?
**Balanced Chemical Reaction:**
\[2 \, \text{AgNO}_3\,(aq) + \text{CaI}_2\,(aq) \rightarrow 2 \, \text{AgI}(s) + \text{Ca(NO}_3\text{)}_2\,(aq)\]
**Calculation Steps:**
1. **Identify the Starting Amount:**
- Given Volume: \( 75.0 \, \text{mL} \)
- Molarity of AgNO₃: \( 0.300 \, \text{M} \)
2. **Calculate Moles of AgNO₃:**
- Convert mL to L: \( 75.0 \, \text{mL} = 0.075 \, \text{L} \)
- Moles of AgNO₃: \( 0.300 \, \text{M} \times 0.075 \, \text{L} = 0.0225 \, \text{moles} \, \text{AgNO₃} \)
3. **Use the Mole Ratio from the Balanced Equation:**
- According to the stoichiometry, \( 2 \) moles of AgNO₃ produce \( 2 \) moles of AgI. The ratio is \( 1:1 \).
4. **Calculate the Moles of AgI Produced:**
- Moles of AgI: \( 0.0225 \, \text{moles} \, \text{AgNO₃} \times \left( \frac{2 \, \text{moles} \, \text{AgI}}{2 \, \text{moles} \, \text{AgNO₃}} \right) = 0.0225 \, \text{moles} \, \text{AgI} \)
**Conclusion:**
When 75.0 mL of 0.300 M AgNO₃ completely reacts, 0.0225 moles of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7e669f8d-7c66-4f3f-a51b-1deb6f0faa05%2F431ee481-f5e3-4aa9-a484-c478b60463ea%2Fmghzmq8.png&w=3840&q=75)
Transcribed Image Text:**Stoichiometry Calculation: How Many Moles of AgI will be Formed?**
**Problem Statement:**
How many moles of AgI will be formed when 75.0 mL of 0.300 M AgNO₃ is completely reacted according to the balanced chemical reaction?
**Balanced Chemical Reaction:**
\[2 \, \text{AgNO}_3\,(aq) + \text{CaI}_2\,(aq) \rightarrow 2 \, \text{AgI}(s) + \text{Ca(NO}_3\text{)}_2\,(aq)\]
**Calculation Steps:**
1. **Identify the Starting Amount:**
- Given Volume: \( 75.0 \, \text{mL} \)
- Molarity of AgNO₃: \( 0.300 \, \text{M} \)
2. **Calculate Moles of AgNO₃:**
- Convert mL to L: \( 75.0 \, \text{mL} = 0.075 \, \text{L} \)
- Moles of AgNO₃: \( 0.300 \, \text{M} \times 0.075 \, \text{L} = 0.0225 \, \text{moles} \, \text{AgNO₃} \)
3. **Use the Mole Ratio from the Balanced Equation:**
- According to the stoichiometry, \( 2 \) moles of AgNO₃ produce \( 2 \) moles of AgI. The ratio is \( 1:1 \).
4. **Calculate the Moles of AgI Produced:**
- Moles of AgI: \( 0.0225 \, \text{moles} \, \text{AgNO₃} \times \left( \frac{2 \, \text{moles} \, \text{AgI}}{2 \, \text{moles} \, \text{AgNO₃}} \right) = 0.0225 \, \text{moles} \, \text{AgI} \)
**Conclusion:**
When 75.0 mL of 0.300 M AgNO₃ completely reacts, 0.0225 moles of
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