How many moles of Agl will be formed when 75.0 mL of 0.300 M AgNO, is completely reacted according to the balanced chemical reaction: 2 AgNO,(aq) + Cal(aq) → 2 Agl(s) + Ca(NO₂)₂(aq)
How many moles of Agl will be formed when 75.0 mL of 0.300 M AgNO, is completely reacted according to the balanced chemical reaction: 2 AgNO,(aq) + Cal(aq) → 2 Agl(s) + Ca(NO₂)₂(aq)
Chemistry
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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
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can you help me and line it exactly like this because i keep doing it and its says its wrong
![**How to Calculate Moles of AgI Formed**
**Problem:**
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:
\[ 2 \text{AgNO}_3(\text{aq}) + \text{CaI}_2(\text{aq}) \rightarrow 2 \text{AgI}(s) + \text{Ca(NO}_3\text{)}_2(\text{aq}) \]
**Solution Steps:**
1. **Starting Amount Conversion**
- Initial Volume: 75.0 mL of AgNO₃
- Convert mL to L: \[ \frac{75.0 \text{ mL AgNO}_3}{1000} = 0.075 \text{ L AgNO}_3 \]
2. **Mole Calculation**
- Use the molarity of AgNO₃ to find moles:
\[ 0.075 \text{ L AgNO}_3 \times 0.300 \text{ mol AgNO}_3/\text{L} = 0.0225 \text{ mol AgNO}_3 \]
3. **Final Calculation**
- According to stoichiometry of the reaction, 2 mol of AgNO₃ forms 2 mol of AgI:
- Therefore, \[ 0.0225 \text{ mol AgNO}_3 \rightarrow 0.0225 \text{ mol AgI} \]
**Graphical Interface Analysis:**
- **Conversion Factors**:
- \( \frac{1 \text{ L AgNO}_3}{1000 \text{ mL AgNO}_3} \)
- \( \frac{0.300 \text{ mol AgNO}_3}{1 \text{ L AgNO}_3} \)
- **Answer**:
- The system provides the answer as \( 0.0225 \text{ mol AgI} \).
- **Interface Tools**:
- Options to multiply or delete steps, enter the answer, reset the calculation, and various numerical and unit inputs to aid in the calculation.
Ensure to verify each step for accuracy and consistency with the stoichiometry of the reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F759a38a4-30b5-475d-b0a8-17aa3311eba3%2Fd312cb0e-421b-4921-aa89-850687a543c5%2F12qkopg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**How to Calculate Moles of AgI Formed**
**Problem:**
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:
\[ 2 \text{AgNO}_3(\text{aq}) + \text{CaI}_2(\text{aq}) \rightarrow 2 \text{AgI}(s) + \text{Ca(NO}_3\text{)}_2(\text{aq}) \]
**Solution Steps:**
1. **Starting Amount Conversion**
- Initial Volume: 75.0 mL of AgNO₃
- Convert mL to L: \[ \frac{75.0 \text{ mL AgNO}_3}{1000} = 0.075 \text{ L AgNO}_3 \]
2. **Mole Calculation**
- Use the molarity of AgNO₃ to find moles:
\[ 0.075 \text{ L AgNO}_3 \times 0.300 \text{ mol AgNO}_3/\text{L} = 0.0225 \text{ mol AgNO}_3 \]
3. **Final Calculation**
- According to stoichiometry of the reaction, 2 mol of AgNO₃ forms 2 mol of AgI:
- Therefore, \[ 0.0225 \text{ mol AgNO}_3 \rightarrow 0.0225 \text{ mol AgI} \]
**Graphical Interface Analysis:**
- **Conversion Factors**:
- \( \frac{1 \text{ L AgNO}_3}{1000 \text{ mL AgNO}_3} \)
- \( \frac{0.300 \text{ mol AgNO}_3}{1 \text{ L AgNO}_3} \)
- **Answer**:
- The system provides the answer as \( 0.0225 \text{ mol AgI} \).
- **Interface Tools**:
- Options to multiply or delete steps, enter the answer, reset the calculation, and various numerical and unit inputs to aid in the calculation.
Ensure to verify each step for accuracy and consistency with the stoichiometry of the reaction.
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